B TON C P NN THII HC 1 I. PHN CHUNG Cu 1: ( 2 im) Cho hm s( )mC m m x m x y 5 5 ) 2 ( 22 2 4+ + + =1, Kho st s bin thin v v th hm s khi m = 1. 2, Vi nhng gi tr no ca m th th ( Cm) c im cc i v im cc tiu, ng thi cc im cc i v im cc tiu lp thnh mt tam gic u. Cu 2: ( 2 im) 1, Gii phng trnh:( )21) 3 cos 1 )( 2 cos 1 ( cos 1 = + + + x x x2, Gii h phng trnh: = + += + + + + + + 1 ) 4 ( log ) 5 ( log6 ) 1 2 ( log ) 2 2 ( log 22 122 1x yx x y x xyy xy x Cu 3: ( 2 im ) 1, Tnh tch phn: ( )}=1314313dxxx xI . 2, Cho cc s thc dng a, b, c tho mnabc ca bc ab = + + . Chng minh rng:

( )1) ( ) (3 34 43 34 43 34 4>++++++++a c caa cc b bcc bb a abb a Cu 4: ( 2 im ) Trong khng gian vi h trc to cc Oxyz, cho mt phng (P) c phng trnh:0 1 2 = + + z y xv ng thng ( d) c phng trnh: = + += 0 2 20 2 2z yy x 1, Tm to giao im A ca ( d) v (P). Tnh s o gc to bi ( d) v (P). 2, Vit phng trnh ng thng( ) A iqua A,( ) Anm trong (P) sao cho gc to bi hai ng thng( ) Av( d) bng 450. II. Phn r i ng ( Th sinh ch lm mt trong hai phn)Cu 5A: ( 2 im ) ( Dnh cho THPT khng phn ban) 1, Vit phng trnh ng trn i qua hai im A( 2;5 ), B9 4; 1) v tip xc vi ng thng c phng trnh:0 9 3 = + y x . 2, Vi n l s nguyn dng, chng minh h thc:( ) ( ) ( )nnnn n nCnC n C C22 22212... 2 = + + +Cu 5B: ( 2 im) ( Dnh cho THPT phn ban) 1, Gii phng trnh:( ) x x x 4 log 1 log41) 3 ( log212842= + + . 2, Cho hnh chp t gi c u S. ABCD c c nh y bng a, chi u cao cng bng a. GiE, Kl n lt l trung im ca cc cnh AD v BC. Tnh bn knh ca mt cu ngoi tip hnh chp S. EBK. P N 1 I. Phn c hung Cu 1: ( 2 im) Cho hm s( )mC m m x m x y 5 5 ) 2 ( 22 2 4+ + + =1, Kho st s bin thin v v th hm s khi m = 1. 2, Vi nhng gi tr no ca m th th ( Cm) c im cc i v im cc tiu, ng thi cc im cc i v im cc tiu lp thnh mt tam gic u. k ( Cm) c 3 im cc tr l m < 2. Cc im cc tr ca ( Cm) l ( ) ( ) ( ) m m C m m B m m A + 1 ; 2 ; 1 ; 2 ; 5 5 ; 02 www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 2 p s: 33 2 = mCu 2: ( 2 im) 1, Gii phng trnh:( )21) 3 cos 1 )( 2 cos 1 ( cos 1 = + + + x x xa phng trnh v dng: 16123cos . cos .2cos2= |.|

\| xxx S dng cng thc bin i tch thnh tng gii hai phng trnh:4123cos . cos .2cos =xxx v 4123cos . cos .2cos =xxx Ta c cc h nghim ca phng trnh cho l:( ) Z m k m xkx e + = + = , 232;2 4 2, Gii h phng trnh: = + += + + + + + + 1 ) 4 ( log ) 5 ( log6 ) 1 2 ( log ) 2 2 ( log 22 122 1x yx x y x xyy xy x K = >= < < 1 ; 20 , 1 4y yx x a phng trnh th nht ca h v dng:( ) 2 1 log ) 2 ( log2 1= + ++ x yy x t) 2 ( log1y tx+ =, tm c t = 1, kt hp vi phng trnh th hai ca h,i chiu vi iu kin trn, tm c nghim( ) ( ) 1 ; 2 ; = y x. Cu 3: ( 2 i m ) 1, Tnh tch phn: ( )}=1314313dxxx xI . a I v dng: }|.|

\| =13133121. 11dxx xI . Dng phng php i bin s, t112 =xt p s: I= 6. 2, Cho cc s thc dng a, b, c tho mnabc ca bc ab = + + . Chng mi nh rng: ( )1) ( ) (3 34 43 34 43 34 4>++++++++a c caa cc b bcc bb a abb a T( ) ( )( ) b a b a ab b b a a b a ab b a b a + + = + + + > + + > +3 3 3 4 3 4 4 4 3 3 4 42 . Vy ( )|.|

\|+ =+>++b a abb ab a abb a 1 12123 34 4. Tng t cho cc bt ng thc cn li, suy ra pcm. Cu 4: ( 2 im ) Trong khng gian vi h trc to cc Oxyz, cho mt phng (P) c phng trnh:0 1 2 = + + z y xv ng thng ( d) c phng trnh: = + += 0 2 20 2 2z yy x 1, Tm to giao im A ca ( d) v (P). Tnh s o gc to bi ( d) v (P). p s. 1)( ) ( )030 ) ( , ; 1 ; 0 ; 1 = Z P d A . 2, Vit phng trnh ng thng( ) A iqua A,( ) Anm trong (P) sao cho gc to bi hai ng thng( ) Av( d) bng 450. Hai ng thng tho mn bi c phng trnh:( ) ( )3 3 513 1 3 21: ;3 3 513 1 3 21:2 1++= = A+=+ =+ Az y x z y x www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 3 II. Phn r i ng ( Th sinh ch lm mt trong hai phn)Cu 5A: ( 2 i m ) ( Dnh cho THPT khng phn ban) 1. Vit phng trnh ng trn iqua haii m A( 2;5 ), B(4; 1) v tip xc vi ng thng c phng trnh:0 9 3 = + y x . Hai ng trn tho mn bi c phng trnh:( ) ( ) ( ) ( ) ( ) ( ) 250 10 17 : ; 10 2 1 :2 222 21= + = + y x C y x C2, Vi n l s nguyn dng, chng minh h thc:( ) ( ) ( )nnnn n nCnC n C C22 22212... 2 = + + +t S l v tri h thc cn chng minh, lu 10= =nn nC Cv k nnknC C=ta thy:( ) ( ) ( ) ( ) ( ) 1 .... 22 212221 nnnn n nC n C n C n C n S + + + + = T( ) ( ) ( ) R x x x xn n ne + = + + , 1 1 12. So s nh hs ca nxtrong khaitri n nhthc Newton ca ( ) ( )n nx x + + 1 1v( )nx21+ta suy ra:( ) ( ) ( ) ( ) 2 ...22 2221 nnnn n nC C C C = + + +T (1) v (2) c pcm. Cu 5B: ( 2 im) ( Dnh cho THPT phn ban) 1, Gii phng trnh:( ) x x x 4 log 1 log41) 3 ( log212842= + + . k x > 0 v1 = x . a phng trnh v dng( ) x x x 4 log 1 log ) 3 ( log2 2 2= + + . Xt hai kh nng 0 < x < 1 v x > 1, i chiu vi iu kin ta tm c hai nghim ca phng trnh l:3 2 3 + = x v x = 3. 2, Cho hnh chp t gi c u S. ABCD c c nh y bng a, chi u cao cng bng a. GiE, K l n lt l trung im ca cc cnh AD v BC. Tnh bn knh ca mt cu ngoi tip hnh chp S. EBK. p s: 829 aR = . 2 Cu 1: Cho hm s y = 2 32xx c th l (C) 1) Kho st s bin thin v v th (C) ca hm s trn. 2) Tm trn (C) nhng im M sao cho tip tuyn ti M ca (C) ct 2 tim cn ca (C) ti A, b sao cho AB ngn nht Cu 2:1/.Gii phng trnh:2 2 si n( ). cos 112x x =2/.Gii h phng trnh: 3 3 32 28 27 18 (1)4 6 (2)xy yxy x y+ = + = Cu 3:1) Tnh tch phn I=2261sin sin2x x dx +} 2) Tm cc gi tr ca tham s thc m sao cho phng trnh sau c nghim thc: (m - 3) x+ ( 2- m)x + 3 - m = 0. (1) Cu 4: Choba s thc dng a, b, c tha mnabc = 1. Chng minh rng:www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 4 3 3 318 1 8 1 8 1a b cc a b+ + >+ + + Cu 5: Cho hnh chp S. ABC c gc ((SBC), (ACB)) =600, ABC vSBC l cc tam gic u cnh a. Tnh theo a khong cch t B n mt phng (SAC).Phn ring: 1.Theo chng trnh chun: Cu6a:ChoAABCcB(1;2),phngictronggcAcphngtrnh(A)2x+y1=0; khong cch t C n (A ) bng 2 ln khong cch t B n (A). Tm A, C bit C thuc trc tung.Cu 7a:Trong khng gian Oxyz cho mp(P): x 2y +z -2 =0 vhai ng thng : (d1) 32 11 1 2yz x+ += = ; (d2) 1 22 ( )1x ty t tz t= + = + e = + . Vit phng trnh tham s ca ng thng Anm trong mp(P) v ct c 2 ng thng (d1) , (d2) 2.Theo chng trnh nng cao: Cu 6b: Cho AABC c din tchbng 3/2;A(2;3), B(3;2), trng tm G e (d) 3xy 8 =0. tm bn kinh ng trn ni tip A ABC. Cu 7b: Trong khng gian Oxyz cho ng thng (d) l giao tuyn ca 2 mt phng:(P): 2x2yz +1 =0, (Q): x+2y 2z 4 =0 v mt cu (S): x2 +y2 +z2 +4x 6y +m =0.Tm tt c cc gi tr ca m (S) ct (d) ti 2 im MN sao cho MN= 8. P N 2 Cu 1: Cho hm s y = 2 32xx c th l (C) 1) Kho st s bin thin v v th (C) ca hm s trn. 2) Tm trn (C) nhng im M sao cho tip tuyn ti M ca (C) ct 2 tim cn ca (C) ti A, B sao cho AB ngn nht GiM(xo; 002 32xx)e (C) . Phng trnh tip tuyn ti M: (A) y =20 02 20 02 6 6( 2) ( 2)x xxx x ++ (A ) TC = A (2; 002 22xx) (A ) TCN = B (2x0 2; 2) 002(2 4; )2AB xx= AB = 20 2044( 2) 2 2( 2)cauchyxx +> AB min =2 2 03 (3; 3)1 (1;1)ox Mx M=

=

Cu 2:1)Gii phng trnh:2 2 si n( ). cos 112x x =phng trnh 2(cosxsinx)(sinx 3 cosx)=0 3( )4x kkx k

= +

e

= +

www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 5 2).Gii h phng trnh: 3 3 32 28 27 18 (1)4 6 (2)xy yxy x y+ = + = (1) y = 0 H 3333222738 18(2 ) 184 63 3 12 . 2 3xxyyx xx xy yy y| |+ =+ =| \ . | | + =+ = | \ . t a = 2x; b = 3y. Ta c h: 3 33 181 ( ) 3a b a bab aba b+ = + = = + = H cho c 2 nghim 3 5 6 3 5 6; , ;4 4 3 5 3 5| | | | + ||\ . \ . + Cu 3:1) Tnh tch phn I=2261sin sin2x x dx +} I=2263cos (cos )2 }x d x. t 3cos cos2x u = I} =242sin23udu=( )3216 +2) Tm cc gi tr ca tham s thc m sao cho phng trnh sau c nghim thc: (m - 3) x+ ( 2- m)x + 3 - m = 0. (1)k x > 0. t t =x ; t > 0 (1)tr thnh (m3)t+(2-m)t2 +3-m = 0 222 3 31t tmt t += +(2)Xt hm s f(t) = 222 3 31t tt t + + (t > 0)Lp bng bin thin(1) c nghim (2) c nghim t > 0 533m s s Cu 4: Choba s thc dng a, b, c tha mnabc = 1. Chng minh rng:3 3 318 1 8 1 8 1a b cc a b+ + >+ + + 3 2 28 1 (2 1)(4 2 1) 2 1cauchyc c c c c + = + + s + 232 18 1a acc>++ Tng t, 2 23 3;2 1 2 18 1 8 1b b c ca ba b> >+ ++ + Ta s chng minh: 2 2 21 (1)2 1 2 1 2 1a b cc a b+ + >+ + + www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 6 Bt(1) 4(a3b2+b3a2+c3a2) +2(a3+b3+c3 )+2(ab2+bc2+ca2)+( a+b+c) >> 8a2b2c2 +4(a2b2 +b2c2 +c2a2) +2 (a2 +b2 +c2 )+1 (2) Ta c:2a3b2 +2ab2 > 4a2b2; .(3) 2(a3b2+b3a2+c3a2) > 2.3.3 5 5 5abc =6 (do abc =1)(4) a3+b3+c3 > 3abc =3 = 1 +2 a2b2c2(5) a3 +a > 2a2; .(6) Cng cc v ca (3), (4), (5), (6), ta c (2).Du bng xy ra khi a=b=c=1 Cu 5: Cho hnh chp S. ABC c gc ((SBC), (ACB)) =600, ABC vSBC l cc tam gic ucnh a. Tnh theo a khong cch t B n mt phng (SAC).Gi M l trung im ca BC v O l hnh chiu ca S ln AM. Suy ra: SM =AM =32a; 060 AMS =v SO mp(ABC) d(S; BAC) = SO =34a V(S.ABC) =33 1( ).3 16adt ABCSO =Mt khc, V(S.ABC) =1( ). ( ; )3dt SACdB SAC ASAC cn ti C c CS =CA =a; SA =32a dt(SAC) = 213 316a Vy d(B; SAC) = 3 3( )13V adt SAC =Phn ring: 1.Theo chng trnh chun: Cu6a:ChoAABCcB(1;2),phngictronggcAcphngtrnh(A)2x+y1=0; khong cch t C n (A ) bng 2 ln khong cch t B n (A). Tm A, C bit C thuc trc tung.Gi H, I ln lt l hnh chiu ca B, C ln (A). M l i xng ca B qua A M e AC v M l trung im ca AC. (BH):x 2y + 3 =0 H( )7 1;5 5 M ( )7 4;5 5 BH = 3 55CI = 6 55; Ce Oy C(0; y0) 075oyy=

= C(0; 7) A ( )27 14;5 5 e(A)loi (0; 5) A( )33 14;5 5e(A) nhn. Cu 7a:Trong khng gian Oxyz cho mp(P): x 2y +z -2 =0 vhai ng thng : (d1) 32 11 1 2yz x+ += = ;(d2) 1 22 ( )1x ty t tz t= + = + e = + .Vitphngtrnhthamscang thng Anm trong mp(P) v ct c 2 ng thng (d1) , (d2) (P) (d1) = A(1;1;2);(P) (d2) = B(3;3;2) (A)1 21 2 ( )2x ty t tz= = e =

2.Theo chng trnh nng cao: Cu 6b: Cho AABC c din tchbng 3/2;A(2;3), B(3;2), trng tm G e (d) 3xy 8 =0. tm bn kinh ng trn ni tip A ABC. www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 7 C(a; b) , (AB): x y 5 =0 d(C; AB) = 5 22ABCa b SABA = 8(1)5 32(2)a ba ba b =

= = Trng tm G ( )5 5;3 3a b + e (d) 3a b =4 (3) (1), (3) C(2; 10) r = 32 65 89Sp =+ + (2), (3) C(1; 1) 32 2 5Srp= =+ Cu 7b: Trong khng gian Oxyz cho ng thng (d) l giao tuyn ca 2 mt phng:(P): 2x2yz +1 =0, (Q): x+2y 2z 4 =0 v mt cu (S): x2 +y2 +z2 +4x 6y +m =0.Tm tt c cc gi tr ca m (S) ct (d) ti 2 im MN sao cho MN= 8. (S) tm I(-2;3;0), bn knh R=13 ( 13) m I Mm = = = m m CC ABTmm bit rng gc gia hai ng thng' ABv' BCbng 060 . Cu V. (1,0 im)Cho cc s thc khng mz y x , ,tho mn32 2 2= + + z y x . Tm gi tr ln nht ca biu thcz y xzx yz xy A+ ++ + + =5. B. PHN RING (3,0 im) Th sinh ch c lm mt trong hai phn (phn a, hoc b). a. Theo chng trnh Chun: Cu VIa. (2,0 im)1.Trong mt phng vi h to , Oxycho tam gicABCc) 6 ; 4 ( A ,phng trnh cc ng thng cha ng cao v trung tuyn k t nhCln lt lwww.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 8 0 13 2 = + y xv0 29 13 6 = + y x . Vit phng trnh ng trn ngoi tip tam gicABC . 2.Trongkhnggianvihto, Oxyz chohnhvungMNPQc) 4 ; 3 ; 2 ( ), 1 ; 3 ; 5 ( P M . Tm to nhQ bit rng nhNnm trong mt phng. 0 6 : ) ( = + z y x Cu VIIa. (1,0 im)Cho tp{ } 6 , 5 , 4 , 3 , 2 , 1 , 0 = E . T cc ch s ca tpElp c bao nhius t nhin chn gm 4 ch s i mt khc nhau? b. Theo chng trnh Nng cao: Cu VIb. 1. Trong mt phng vi h to , Oxyxt elp) (Ei qua im) 3 ; 2 ( Mv cphng trnh mt ng chun l. 0 8 = + xVit phng trnh chnh tc ca ). (E2.Trongkhnggianvihto, Oxyz choccim) 2 ; 3 ; 0 ( ), 0 ; 1 ; 0 ( ), 0 ; 0 ; 1 ( C B A vmt phng. 0 2 2 : ) ( = + + y x Tm to ca imMbit rngMcch u cc imC B A , , v mt phng). (Cu VIIb. (1,0 im)Khai trin v rt gn biu thc nx n x x ) 1 ( ... ) 1 ( 2 12 + + + thu c a thc nnx a x a a x P + + + = ... ) (1 0. Tnh h s 8abit rngnl s nguyn dng tho mn n C Cn n1 7 13 2= + . P N 3 A. PHN CHUNG CHO TT C TH SINH (7,0 im) Cu I. (2,0 im)Cho hm sm x x m x y + + = 9 ) 1 ( 32 3, vim l tham s thc. 1.Kho st s bin thin v v th ca hm s cho ng vi1 = m . V i 1 = mta c1 9 62 3 + = x x x y . *Tp x c nh: D = R *S bi n thi n- Chi u bi n thi n:) 3 4 ( 3 9 12 3 '2 2+ = + = x x x x yTa c

>130 'xxy ,3 1 0 ' < < < x y . Do : + Hm s ng bin trn mi khong) 1 , (v) , 3 ( + . + Hm s nghch bi n tr n khong ). 3 , 1 (-Cctr:Hmstcciti 1 = x v3 ) 1 ( = = y yCD; tccti ut i 3 = x v 1 ) 3 ( = = y yCT. -Gi ih n:+ = =+ y yx xlim ; lim . - Bng bi n thi n: - th: thct trc tung t ii m) 1 , 0 ( . 2.Xc nhm hm s cho t cc tr ti 2 1, x xsao cho22 1s x x . Ta c. 9 ) 1 ( 6 3 '2+ + = x m x y+) Hm s t cc i, cc tiu ti 2 1, x x www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 9 phng trnh0 ' = yc hai nghim pb l2 1, x x Pt0 3 ) 1 ( 22= + + x m xc hai nghim phn bit l2 1, x x .

> + = A 3 13 10 3 ) 1 ( '2mmm ) 1 (+) Theo nh l Vi et ta c. 3 ); 1 ( 22 1 2 1= + = + x x m x xKhi ( ) ( ) 4 12 1 4 4 4 222 122 1 2 1s + s + s m x x x x x x ) 2 ( 1 3 4 ) 1 (2s s s + m mT (1) v (2) suy ra gi tr ca m l3 1 3 < s mv. 1 3 1 s < + mCu II. (2,0 im) 1.Gii phng trnh: )2sin( 2cos sin2 sincot21 + =++ xx xxx . i u ki n:. 0 cos sin , 0 sin = + = x x xPt cho tr thnh 0 cos 2cos sincos sin 2sin 2cos= ++ xx xx xxx 0 2 sin )4sin( cos0cos sincos 2sin 2cos2= |.|

\| + =+ x x xx xxxx

+). ,20 cos Z e + = = k k x x +)Z e

+ =+ =

+ =+ + = + = n mnxm xn x xm x xx x ,32424242242)4sin( 2 sin . ,324Z e + = ttx i chiu iu kin ta c nghim ca pt lk x + =2;. , ,324Z e + = t ktx 2.Gii phng trnh: ) 1 2 ( log 1 ) 1 3 ( log 2355+ = + x x . i u ki n.31> x(* ) V ik tr n, pt cho) 1 2 ( log 3 1 ) 1 3 ( log525+ = + x x

3 23525) 1 2 ( ) 1 3 ( 5) 1 2 ( log ) 1 3 ( 5 log+ = + = x xx x

www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 10

=== = + 8120 ) 1 8 ( ) 2 (0 4 36 33 822 3xxx xx x x i chiu iu kin (*), ta c nghim ca pt l. 2 = x Cu III. (1,0 im)Tnh tch phn}++=5121 31dxx xxI . t 321 3 231 3tdtdxxdxdt x t = += + = .Khi 1 = xth t = 2, v khi x = 5 th t = 4. Suy ra }+||.|

\| =4222232..31131tdttttI} }+ =42242212 ) 1 (92tdtdt t.59ln271002411ln2431923+ =++|.|

\| =ttt tCu IV. (1,0 im)Cho hnh lng tr tam gic u' ' ' . C B A ABCc). 0 ( ' , 1 > = = m m CC ABTmm bit rng gc gia hai ng thng' ABv' BCbng 060 . - K) ' ' ( ' // B A D AB BD e060 ) ' , ( ) ' , ' ( = = BC BD BC AB

060 ' = Z DBChoc. 120 '0= ZDBC- Nu 060 ' = ZDBCVl ng tr u n n). ' ' ' ( ' C B A BB p dng nh l Pitago v nh l cosin ta c 1 '2+ = = m BC BDv. 3 ' = DC Kt h p 060 ' = ZDBCta suy ra' BDC Au.Do . 2 3 12= = + m m- Nu 0120 ' = ZDBC p dng nh l cosi n cho' BDC A suy ra0 = m(l o i ). Vy. 2 = m*Ch :- Nu HS ch xt trng hp gc 060thchcho 0,5 khigi ing. - HS c th gii bng phng php vect hoc to vi nhn xt:' '.' . ') ' , ' cos( ) ' , ' cos(BC ABBC ABBC AB BC AB = = . A C C B B A m D 31 1 0120www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 11Cu V. (1,0 im)Cho cc s thc khng mz y x , ,tho mn32 2 2= + + z y x . Tm gi tr ln nht ca biu thcz y xzx yz xy A+ ++ + + =5. tz y x t + + = 23) ( 2 322= + + + + + =tzx yz xy zx yz xy t . Ta c3 02 2 2= + + s + + s z y x zx yz xyn n3 3 9 32s s s s t tv . 0 > tKhi.5232ttA +=Xt hm s. 3 3 ,23 52) (2s s + = tttt fTa c05 5) ( '232>= =tttt t fv . 3 > tSuy ra) (t fng bi n tr n] 3 , 3 [ . Do .314) 3 ( ) ( = sf t fDu ng thc xy ra khi . 1 3 = = = = z y x tVy GTLN ca A l 314, t c khi. 1 = = = z y xB. PHN RING (3,0 im) Th sinh ch c lm mt trong hai phn (phn a, hoc b). a. Theo chng trnh Chun: Cu VIa. (2,0 im)1. Trong mt phng vi h to , Oxycho tam gicABCc ) 6 ; 4 ( A ,phng trnh cc ng thng cha ng cao v trung tuyn k t nhClnlt l0 13 2 = + y xv0 29 13 6 = + y x . Vit phng trnh ng trn ngoi tip tamgicABC . - Gi ng cao v trung tuyn kt C l CH v CM. Khi CH c phng trnh0 13 2 = + y x , CM c phng trnh. 0 29 13 6 = + y x- T h ). 1 ; 7 (0 29 13 60 13 2 = + = + Cy xy x - ) 2 , 1 ( = = CH ABu n CH AB 0 16 2 : = + y x AB pt . - T h ) 5 ; 6 (0 29 13 60 16 2My xy x= + = +

). 4 ; 8 ( B - Gi s phng trnh ng trn ngoi tip. 0 :2 2= + + + + A p ny mx y x ABCVA, B, C thuc ng trn nn = + = + + += + + +0 7 500 4 8 800 6 4 52p n mp n mp n m == =7264pnm. Suy ra pt ng trn:0 72 6 42 2= + + y x y xhay. 85 ) 3 ( ) 2 (2 2= + + y xM(6; 5) A(4; 6) C(-7; -1) B(8; 4) H www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 122.Trongkhnggianvihto, Oxyz chohnhvungMNPQc) 4 ; 3 ; 2 ( ), 1 ; 3 ; 5 ( P M . Tm to nhQ bit rng nhNnm trong mt phng. 0 6 : ) ( = + z y x - Gi s) ; ; (0 0 0z y x N . V ) 1 ( 0 6 ) (0 0 0= + e z y x N - MNPQ l hnh vungMNP A vung cn t iN ==0 .PN MNPN MN

= + + + + + + + = + + + 0 ) 4 )( 1 ( ) 3 ( ) 2 )( 5 () 4 ( ) 3 ( ) 2 ( ) 1 ( ) 3 ( ) 5 (0 020 0 0202020202020z z y x xz y x z y x = + + + + = +) 3 ( 0 ) 4 )( 1 ( ) 3 ( ) 2 )( 5 () 2 ( 0 10 020 0 00 0z z y x xz x - T (1) v (2) suy ra + =+ =17 20 00 0x zx y. Thay vo (3) ta c0 6 5020= + x x

= = = = = =2 , 1 , 31 , 3 , 20 0 00 0 0z y xz y x hay

) 2 ; 1 ; 3 () 1 ; 3 ; 2 (NN. - GiIl tm hnh vung Il trung imMP v N Q )25; 3 ;27( I . Nu) 1 3 ; 2 ( Nth ). 4 ; 3 ; 5 ( QNu) 2 ; 1 ; 3 ( Nth ). 3 ; 5 ; 4 ( QCu VIIa. (1,0 im)Cho tp{ } 6 , 5 , 4 , 3 , 2 , 1 , 0 = E . T cc ch s ca tpElp c bao nhiu st nhin chn gm 4 ch s i mt khc nhau? Gi sabcdl s tho mn ycbt. Suy ra{ } 6 , 4 , 2 , 0 e d . +). 0 = dS c ch sp xpabcl.36A+). 2 = dS c ch sp xpabcl.2536A A +) V i4 = dhoc6 = dkt qu ging nh trng hp. 2 = dDo ta c s cc s lp c l( ) . 420 3253636= + A A Ab. Theo chng trnh Nng cao: CuVIb.(2,0im)1.Trongmtphngvihto, Oxy xtelp) (E iquaim ) 3 ; 2 ( M vcphngtrnhmtngchunl. 0 8 = + x Vitphngtrnhchnhtc ca). (E- Gi phng trnh) 0 ( 1 : ) (2222> > = + b abyaxE . - Gi thi t== +) 2 ( 8) 1 ( 19 422 2cab a Ta c ). 8 ( 8 8 ) 2 (2 2 2 2 2c c c c c a b c a = = = = Thay vo (1) ta c1) 8 (984=+c c c. www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 13

== = + 21320 26 17 22ccc c*Nu2 = cth . 112 16: ) ( 12 , 162 22 2= + = =y xE b a*Nu 213= cth . 14 / 39 52: ) (439, 522 22 2= + = =y xE b a2.Trongkhnggianvihto, Oxyz choccim) 2 ; 3 ; 0 ( ), 0 ; 1 ; 0 ( ), 0 ; 0 ; 1 ( C B A vmt phng. 0 2 2 : ) ( = + + y x Tm to ca imMbit rngMcch u cc imC B A , , v mt phng). (Gi s) ; ; (0 0 0z y x M .Khi t gi thi t suy ra 52 2) 2 ( ) 3 ( ) 1 ( ) 1 (0 0 202020202020202020+ += + + = + + = + + y xz y x z y x z y x+ += + + + + = + ++ + = + + ) 3 (5) 2 2 () 1 () 2 ( ) 2 ( ) 3 ( ) 1 () 1 ( ) 1 ( ) 1 (20 0 202020202020202020202020202020y xz y xz y x z y xz y x z y x T (1) v (2) suy ra ==0 00 03 x zx y.Thay vo (3) ta c20 020) 2 3 ( ) 10 8 3 ( 5 + = + x x x

==323100xx

).314;323;323() 2 ; 1 ; 1 (MM Cu VIIb. (1,0 im)Khai trin v rt gn biu thc nx n x x ) 1 ( ... ) 1 ( 2 12 + + + thu c a thc nnx a x a a x P + + + = ... ) (1 0. Tnh h s 8abit rngnl s nguyn dng tho mn n C Cn n1 7 13 2= + . Ta c= +> = +n n n n n nnn C Cn n1) 2 )( 1 (! 3 . 7) 1 (231 7 13 2 . 90 36 532= = > nn nn Suy ra 8al h s ca 8xtrong bi u thc. ) 1 ( 9 ) 1 ( 89 8x x + l . 89 . 9 . 88988= + C C 4 www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 14I:PHN CHUNG CHO TT C CC TH SINH (7,0 im) Cu I. (2,0 im)Cho hm s 1 x2 xy+= (C) 1. Kho s t s bin thin v v th hm s (C). 2. Cho im A(0;a) .Xc nh a t A k c hai tip tuyn ti (C) sao chohai tip im tng ng nm v hai pha trc ox. Cu II. (2,0im) 1. Gii h phng trnh : = + += + + 0 22 20 9 6 42 22 2 4y x y xy y x x. 2. Gii PT:( )2 22 1cos cos sin +13 3 2x x x | | | |+ + + = ||\ . \ . Cu III. (1,0im)Tnh tch phn I=6 644sin cos6 1xx xdx++} CuIV.(2,0im)ChohnhchpS.ABCD,c yABCDlhnhvungtmOc nhbnga, SO (ABCD). GiM, Nln lt l trung im ca SA v BC. Tnh gc gia ng thng MNv mt phng (ABCD) v thtch khichp M.ABCD, bi t rng.210 aMN =Cu V (1 i m) Cho ba s a, b, c sao cho =>10 , ,abcc b a.Tm gi trnh nht ca bi u thcA = ( )++ c b a31( )++ c a b31( ) a b c +31 Phn Ri ng: (3 i m) Thsi nh ch c chn lm mt trong hai phn (phn A hoc phn B) A. Theo chng trnh chun. Cu VI .a (2 i m) 1)ChoAABC c PT hai cnh l: 0. 21 - 7y 4x = + = + , 0 6 2 5 y x Trc tm ca tam gic trng vigc to O, lp phng trnh cnh cn li. 2.Trong khng gian vi h ta Oxyz, cho im M(2 ; 1 ; 0) v ng thng d v id :x 1 y 1 z2 1 1 += =.Vit phng trnh chnh tc ca ng thng i qua im M,ct v vung gc vi ng thng d v tm to ca im M i xng vi M qua d CuVI I .a (1i m)Mt lp hc c 40 hc sinh, cn c ra mt ban cn s gm mt lp trng, mtlpphv3yvin(Bitrngkhngphn bitccchcdanhlyvin).Hicbao nhi u c ch l p ra mt ban c n s. B. Theo chng trnh nng cao. Cu VI .b (2 i m) 1)Trong mt phng vi h trc ta Oxy cho A(4;3), ng thng (d) :x y 2 = 0 v (d): x + y 4 = 0 ct nhau ti M. Tm( ) ( ') B d v C d e esao cho A l tm ng trn ngoi tip tam gic MBC. 2) Trong kg Oxyz cho ng thng ( A): x= -t ; y=2t -1 ; z=t +2v mp(P):2x y -2z - 2=0 Vit PT mt cu(S) c tm IeAv khong cch t I n mp(P) l 2 v mt cu(S) ct mp(P )theo giao tuyn ngtrn (C)c bn knh r=3Cu VI I .b (1 i m) Tm c c gi trca tham s m ng thngm x y + = 2ct th hm s xx xy12 +=t ihaii m phn bi t A, B sao cho trung i m ca o n thng AB thuc trc tung. www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 15 P N 4 I:PHN CHUNG CHO TT C CC TH SINH (7,0 im) Cu I. (2,0 im)Cho hm s 1 x2 xy+= (C) 1. Kho st s bin thin v v th hm s (C). 2. Cho i m A(0;a) .Xc nh a t A k c hai tip tuyn ti (C) sao cho hai tip im tng ng nm v hai pha trc ox.Phng trnh tip tuyn qua A(0;a) c dng y=kx+a (1) i u ki n c haiti p tuyn qua A:= =+) 3 ( k) 1 x (3) 2 ( a kx1 x2 x2 c nghi m1 x =Thay (3) vo (2) v rt gn ta c: ) 4 ( 0 2 a x ) 2 a ( 2 x ) 1 a (2= + + + (4) c 2 nghi m1 x =l: >=> + = A= ==2 a1 a0 6 a 3 '0 3 ) 1 ( f1 a Honh tip im 2 1x ; xl nghim ca (4)Tung tip im l1 x2 xy111+= , 1 x2 xy222+= hai tip im nm v hai pha ca trc ox l: 0) 2 x )( 1 x () 2 x )( 2 x (0 y . y2 12 12 1< + + = ++ += +} }}t tt tt t t tI dt I dtt tdt

424 44 443 5 3 5 3 12 1 sin 2 cos 4 sin 44 8 8 8 8 45 516 32 | | | | | |= = + = + |||\ . \ . \ .= =} }I t dt t dt t tI CuIV.(2,0im)ChohnhchpS.ABCD,c yABCDlhnhvungtmOc nhbnga, SO (ABCD). GiM, Nln lt l trung im ca SA v BC. Tnh gc gia ng thng MNv mt phng (ABCD) v thtch khichp M.ABCD, bi t rng.210 aMN =SO (ABCD). Dng MH//SO, H thuc AC, khi MH (ABCD), suy ra gc gi a ng thng MN v imp(ABCD) chnh l gc. = H N MTa cn tnh . Xt tam gi c CN H c :.2,42 3.43 aCNaAC HC = = =0 2 2 245 cos . . 2 CN HC CN HC HN + =Hay 434 892 2 22a a aHN + =C A S B D O NH M a 210 a www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 17Suy ra.410 aHN =Vy 21102.410cos = = =aaMNHN . Dn n. 600= Vy gc gia ng thng MNv mt phng (ABCD) bng 600. Thtch khichp M.ABCD. Trong tam gi c HMNc,

83023.41060 tan . 60 tan0 0a aHN MHHNMH= = = = . MH l chiu cao ca khi chp M.ABCD. Vy th tch ca khi chp ny l: .2430830.31.3132a aa MH S VABCD= = = Cu V (1 i m) Cho ba s a, b, c sao cho =>10 , ,abcc b a.Tm gi trnh nht ca bi u thcA = ( )++ c b a31( )++ c a b31( ) a b c +31 tx = czbya1,1,1= = .Khi: =+++++=x yzz xyz yxA1 1 1 1 1 13 3 3233 3 3>+++++ y xxy zx zxz yz yyz x(* ) Do1 1 = = xyz abc n n ta c y xzx zyz yxA+++++=2 2 2(1) Ta chng mi nh bt ng thc 2c b a + +.2 2 2a bca cbc ba+++++s Tht vy.p dng bt ng thc Csi cho cc s dng ta c: ac bc ba>+++ 42,ba ca cb>+++ 42,cb ab ac>+++ 42. Cng ba bt ng thc c ng chi u tr n ta c :2c b a + +.2 2 2a bca cbc ba+++++sB n c t nh gi du =xy ra khia = b = c. Vy A=2323232 2 2= >+ +>+++++xyzz y xy xzx zyz yx Du =xy ra khix = y = z = 1. Vy mi nA = 23 khia = b = c =1. Phn Ri ng: (3 i m) Thsi nh ch c chn lm mt trong hai phn (phn A hoc phn B) A. Theo chng trnh chun. Cu VI .a (2 i m) 1)ChoAABC c PT hai cnh l: 0. 21 - 7y 4x = + = + , 0 6 2 5 y x Trc tm ca tam gic trng vigc to O, lp phng trnh cnh cn li. Ta gi s tam gi c ABC c c nh AB :0 6 2 5 = + y xwww.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 18AC:0 21 - 7y 4x = + , suy ra ta ca A l nghim ca h phng trnh: = + = 21 7 46 2 5y xy x,gi ihsuy ra A(0; 3) Nhn thy A thuc Oy, OA l ngcao ca tam gi c,Ox BC BC OA // suy ra phng trnh ca BC c dng y = y0. ng cao BB i qua trc tm O v vung gc vi AC suy ra BB c phng trnh l: 7(x 0) - 4(y 0) = 0 hay BB : 7x 4y = 0. i m B = AC BB' ta ca B l nghim ca h phng trnh: = = = = 746 2 50 4 7yxy xy x ng thng i qua B(- 4; - 7) v song song vi Ox chnh l ng thng BC suy ra phng trnh c nh BC: y = - 7. Vy phng trnh cnh cn li ca tam gic ABC l y = -7. 2.Trong khng gian vi h ta Oxyz, cho im M(2 ; 1 ; 0) v ng thng d v id :x 1 y 1 z2 1 1 += =.Vit phng trnh chnh tc ca ng thng i qua im M, ct v vung gc vi ng thng d v tm to ca im M i xng vi M qua d Gi H l hnh chiu vung gc ca M trn d, ta c MH l ng thng i qua M, ct v vung gc vi d. d c phng trnh tham s l: x 1 2ty 1 tz t= + = + = V H e d nn ta H (1 + 2t ; 1 + t ; t).Suy ra : MH= (2t 1 ; 2 + t ; t) V MH d v d c mt vect ch phng lu = (2 ; 1 ; 1), nn : 2.(2t 1) + 1.( 2 + t) + ( 1).(t) = 0 t = 23.V th,MH = 1 4 2; ;3 3 3| | |\ . 3 (1; 4; 2)MHu MH = = Suy ra, phng trnh chnh tc ca ng thng MH l: x 2 y 1 z1 4 2 = = Theo tr n c 7 1 2( ; ; )3 3 3H m H l trung im ca MM nn to M8 5 4( ; ; )3 3 3 Cu VI I .a (1 i m) Mt l p hc c 40 hc si nh, cn c ra mt ban c n s gm mt l ptrng, mt lp ph v 3 y vin (Bit rng khng phn bit cc chc danh l y vin). Hic bao nhi u c ch l p ra mt ban c n s. uti ntachn ra 2 hc sinh lm lp trng v lp ph, (ch rng hai chc danh l kh c nhau) Mt cch xp 2 hc sinh lm lp trng v lp ph l mt chnh hp chp 2 ca 40 S cch xp 2 hc sinh lm lp trng v lp ph l 240ACn l i38 hc si nh. Tip ta chn 3 hc sinh lm y vin (khng phn bit th t) S cch chn 3 hc sinh lm y vin l 338C Theo qui tc nhn ta c s cch chn ra mt ban cn s l : 13160160 .338240= C Ac ch A B C O(0; 0) A B A www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 19B. Theo chng trnh nng cao. Cu VI .b (2 i m) 1)Trong mt phng vi h trc ta Oxy cho A(4;3), ng thng (d) :x y 2 = 0 v (d): x + y 4 = 0 ct nhau ti M. Tm( ) ( ') B d v C d e esao choA l tm ng trn ngoi tip tam gic MBC. 2) Trong kg Oxyz cho ng thng ( A): x= -t ; y=2t -1 ; z=t +2v mp(P):2x y -2z - 2=0 Vit PT mt cu(S) c tm IeAv khong cch t I n mp(P) l 2 v mt cu(S) ctmp(P )theo giao tuyn ngtrn (C)c bn knh r=3m cu(S) c tm IeAg sI(a;b;c ) =>(a;b;c) tho mn PT ca A(1) * ( ) ( ); 2 d I P =(2) T (1) v(2) ta c h PT:2 2 2 611 14 1 1 1 7.... ; ; ; ; ;2 16 3 6 3 3 32a b ca theconghiem vab tc t = | | | |= ||= \ . \ .= + Do24 3 13 r R R = = =Vy c 2 mt cu theo ycbt : ( )2 2 212 2 2211 14 1( ) : 136 3 61 1 7: 133 3 3S x y zS x y z| | | | | | + + + = |||\ . \ . \ .| | | | | |+ + + + = |||\ . \ . \ . Cu VI I .b (1 i m) Tm c c gi trca tham s m ng thngm x y + = 2ct th hm s xx xy12 +=t ihaii m phn bi t A, B sao cho trung i m ca o n thng AB thuc trc tung. Phng trnh honh giao im:) 0 ( 0 1 ) 1 ( 3 2122= = + + =+ +x x m x m xxx x(1) Nhn thy x =0, khng l nghim ca phng trnh (1) v c bit s: ( ) m m > + = A , 0 12 12, suy ra phng trnh (1) lun c hai phn bit 2 1, x xkh c 0 v imim, tc thng lun ct ng cong ti hai im A, B phn bi t v imim. Theo nh l Vi t ta c 312 1= = +mabx x Honh trung im Ica o n thng AB l 6122 1=+=m x xxI. i m. 1 0 1 0 = = = e m m x Oy II Vy m= 1 l gi tr cn tm. 5 I: PHN CHUNG CHO TT C TH SINH CuI (2im): Cho hm s y = x3 - 3x2 + 4(C) 1: Kho st hm s. 2: Gi (d) l ng thng i qua im A(2 ; 0) c h s gc k.Tm k (d) ctwww.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 20 (C) ti ba im phn bit A ; M ; N sao cho hai tip tuyn ca (C ) ti M v N vung gc vi nhau. Cu II (2 im): 1: Gii phng trnh: x xx x2 si n21cos 2 )2cos2(si n 33 3+ = 2: Gii bt phng trnh:2 235 5 4 24 x x x + < + +Cu III (1im): Tnh tch phn : I = 52ln( 1 1)1 1xdxx x + + } Cu IV (1im):Chotam gic ABC cn ni tip ng trn tm J bn knh R=2a (a>0) ,gc BAC =1200.Trn ng thng vung gc vi mt phng (ABC) l y i m S sao cho SA = 3. a GiI l trung im on BC .Tnh gc gia SI v hnh chiu ca n trn mt phng (ABC) & tnh bn knh mt cu ngoitip hnh chp SABC theo aCu V (1im):1). Cho x,y,z l cc s thc dng. Chng minh: 3 2 3 2 3 2 2 2 22 2 2 1 1 1 y x zx y y z z x x y z+ + s + ++ + + 2).Tm m h phng trnh: 3 3 22 2 23 3 2 01 3 2 0x y y xx x y y m + =+ + =c nghim thc PHN RING: Th sinh ch c lm mt trong hai phn A hoc B A.Theo chng trnh chun (2im) CuVIa:1)TrongmtphngOxychongtrn(C)tmI(-1;1),bnknhR=1,Mlmt im trn( ) : 2 0 d x y + = . Hai tip tuyn qua M to vi (d) mt gc 450 tip xc vi(C) ti A, B. Vit phng trnh ng thng AB. 2) Cho hnh lp phng ABCDA1B1C1D1 c im A(0;0;0); B(2;0;0); D(0;2;0);A1(0;0;2). M l trung im AB; N l tm ca hnh vung ADD1A1. Tnh bn knh ca ng trn l giao tuyn ca mt cu i qua C ; D1 ; M ; N vi mt phngMNC1 Cu VII/a:Cho n l s t nhin n>2.Tnh 2 2 1 2 2 2 212 1 . .2 2 . .2 ... . .2nk k n nn n n nkS kC C C n C== = + + + B. Theo chng trnh nng cao (2im) Cu VIa.1) Cho (P) y2 = x v ng thng (d): x y 2 = 0 ct (P) ti hai im A v B.Tm im C thuc cung AB sao choAABC c din tch ln nht2) Trong khng gian vi h ta Oxyz cho hai ng thng : 11 1( ) :2 1 1x y zd += = v

22 1( ) :1 1 1x y zd += =. Vit phng trnh mtphng cha (d1) v hp vi(d2) mt gc 300. Cu VII/b: Gii h phng trnh 20102 22( 1)log2 3xy xyy x x y = + + = P N 5 www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 21 I: PHN CHUNG CHO TT C TH SINH CuI (2im): Cho hm s y = x3 - 3x2 + 4(C) 1: Kho st hm s. 2: Gi (d) l ng thng i qua im A(2 ; 0) c h s gc k.Tm k (d) ct(C) ti ba im phn bit A ; M ; N sao cho hai tip tuyn ca (C ) ti M v N vung gc vi nhau. +PT ng thng d: y=k(x-2) +Honh A;M;N l nghim PT: x3-3x2+4=k(x-2) (x-2)(x2-x-2-k)=0x=2=xA;f(x)=x2-x-2-k=0 +PT c 3nghim phn bit f(x)=0 c 2nghim phn bit khc 2 090(2) 0 4kfA > < ==.Theo Vit ta c 12M NM Nx xx x k+ = = +Tip tuyn ti M v N vung gc vi nhau y'(xM).y'(xN)=-1 (2 23 6 )(3 6 ) 1M M N Nx x x x = 9k2+18k+1=03 2 23k =(tm) Cu II (2 im): 1: Gii phng trnh: x xx x2 si n21cos 2 )2cos2(si n 33 3+ = PT tng ngx 2 si n21x cos 2 )2xcos2x(si n 33 3+ = ( ) x cos x si n 22xcos2xsi n 12xcos2xsi n 3 + = |.|

\| + |.|

\| ( ) |.|

\|+ |.|

\| + = |.|

\| + |.|

\| 2xsi n2xcos2xsi n2xcos x si n 2 x si n2112xcos2xsi n 30232xcos2xsi n ) x si n 2 (2xsi n2xcos = |.|

\|+ + + |.|

\| * x x x xsi n cos 0 si n 0 k x k2 ( k )2 2 2 4 2 4 2t t t | | = = = t = + t e |\ .Z*2 x si n 0 x si n 2 = = +(v nghi m) *2 234x si n234 2xsi n 2232xcos2xsi n = |.|

\| t+ = |.|

\| t+ = + (v nghi m) Vy nghi m ca phng trnh l:( ) x k2 k2t= + t eZ 2: Gii bt phng trnh:2 235 5 4 24 x x x + < + +BPT tng ng 2 22 22 235 24 5 4115 435 2411 (5 4)( 35 24)x x xxx xx x x+ + < < + + + < + + + Xt: a)Nu x45s khng tha mn BPT b)Nu x>4/5: Hm s 2 2(5 4)( 35 24) y x x x = + + + vi x>4/5 www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 22y'=2 22 21 15( 35 24) (5 4)( )35 24x x xx x+ + + + ++ +>0 mi x>4/5 Vy HSB.+Nu 4/51 th y(x)>11 Vy nghim BPT x>1 Cu III (1im): Tnh tch phn : I = 52ln( 1 1)1 1xdxx x + + } t t= 1 1 x +* x = 2t = 2 *x = 5t = 3*dx=2(t-1)dt I=23 322 232 3 2 222( 1) ln ln2( 1) 12 ln ln ln ln 3 ln 2t t tdt dtt t ttd t t= + = = = } }} Cu IV (1im):Chotam gic ABC cn ni tip ng trn tm J bn knh R=2a (a>0) ,gc BAC =1200.Trn ng thng vung gc vi mt phng (ABC) l y i m S sao cho SA = 3. a Gi I l trung im on BC .Tnh gc gia SI v hnh chiu ca n trn mt phng (ABC) & tnh bn knh mt cu ngoitip hnh chp SABC theo a+Gi D l trung im BCADBC (V ABC cn ti A) AD(SBC) +Gi E trung im SBAE SB (V SAB u) DESB (nh l 3 ng vung gc) +SC//DE (DE ng trung bnh tam gic) SCSB Vy tam gic SBC vung ti S +AD l trc ng trn ngoi tip tam gic SBC.Nn tm O mt cu ngoi tip SABC thuc AD.Mt khc O cch u A; B; C nn O l tm ng trn ngoi tip tam gic ABC.Vy bn knh R mt cu ngoi tip hnh chp bng bn knh ng trn ngoi tip tam gic ABC +BC =2 2a b + 2 2 2 2DC 3cosC= sinAC 2 2a b a bCa a+ = =+ R = 22 22sin3AB aCa b= Cu V (1im):1). Cho x,y,z l cc s thc dng. Chng minh: 3 2 3 2 3 2 2 2 22 2 2 1 1 1 y x zx y y z z x x y z+ + s + ++ + + +t0; 0; 0 a x b y c z = > = > = >+VT=6 4 6 4 6 4 3 2 3 2 3 2 2 2 2 2 2 22 2 2 2 2 2 1 1 12 2 2a b c a b ca b b c c a a b b c c a a b b c c a+ + s + + = + ++ + + (Theo BT CSi) +VP=4 4 4 2 2 2 2 2 21 1 1 1 1 1a b c a b b c c a+ + > + +BCASDEwww.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 23(p dng BT CSi cho tng cp) PCM. Du bng xy ra khi v ch khi a = b = c =1 hay x = y = z = 1 2).Tm m h phng trnh: 3 3 22 2 23 3 2 0 (1)1 3 2 0 (2)x y y xx x y y m + =+ + =c nghim thc iu kin: 221 0 1 10 22 0x xyy y > s s s s > t t = x + 1 te[0; 2]; ta c (1) t3 3t2 = y3 3y2. Hm s f(u) = u3 3u2 nghch bin trn on [0; 2] nn:(1) t = y y = x + 1 (2) 2 22 1 0 x x m + =t 21 v x = ve[0; 1] (2) v2 + 2v 1 = m. Hm s g(v) = v2 + 2v 1 t 0;1 0;1min ( ) 1; m ( ) 2[ ] [ ] ax gv gv = =Vy h phng trnh c nghim khi v ch khi 1 s ms 2 PHN RING: Th sinh ch c lm mt trong hai phn A hoc B A.Theo chng trnh chun (2im) CuVIa:1)TrongmtphngOxychongtrn(C)tmI(-1;1),bnknhR=1,Mlmt im trn( ) : 2 0 d x y + = . Hai tip tuyn qua M to vi (d) mt gc 450 tip xc vi(C) ti A, B. Vit phng trnh ng thng AB. D thy( ) I d e . Hai tip tuyn hp vi (d) mt gc 450 suy ra tam gic MAB vung cn v tam gic IAM cng vung cn . Suy ra:2 IM = . ( ) ( M d M e a; a+2),( 1; 1) IM a a = + +, 02 2 1 22aIM aa= = + = = .Suy ra c 2 im tha mn: M1(0; 2) v M2 (-2; 0). + ng trn tm M1 bn kinh R1=1 l (C1): 2 24 3 0 x y y + + = . Khi AB i qua giao im ca (C ) v (C1) nn AB: 2 2 2 24 3 2 2 1 1 0 x y y x y x y x y + + = + + + + = . + ng trn tm M2 bn kinh R2=1 l (C2): 2 24 3 0 x y x + + + = . Khi AB i qua giao im ca (C ) v (C2) nn AB: 2 2 2 24 3 2 2 1 1 0 x y x x y x y x y + + + = + + + + + = . + KL: Vy c hai ng thng tha mn:1 0 x y + =v1 0 x y + + =2).Cho hnh lp phng ABCDA1B1C1D1 c im A(0;0;0); B(2;0;0); D(0;2;0);A1(0;0;2). M l trung im AB; N l tm ca hnh vung ADD1A1. Tnh bn knh ca ng trn l giao tuyn ca mt cu i qua C ; D1 ; M ; N vi mt phngMNC1 +Mt cu i qua C(2; 2; 0);D1(0; 2; 2);M(1; 0; 0);N(0; 1; 1) c phng trnh: x2+y2+z2+2ax+2by+2cz+d=0 nn www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 244 4 8 04 4 8 05 1; ; 42 1 0 2 22 2 2 0a b db c da c b da db c d+ + + = + + + == = = =+ + =+ + + = Suy ra tm mt cu v bn knh mt cu l: I(5/2;1/2;5/2); R = 352 +(MNC1) i qua M(1;0;0) nhn 1 1; (0;3; 3) MC NC ( = lm vc t php tuyn c PT: y z = 0 + h = d(I;(MNC1)) = 2+ Bn knh ng trn giao tuyn l 2 23 32R h =Cu VII/a:Cho n l s t nhin n>2.Tnh 2 2 1 2 2 2 212 1 . .2 2 . .2 ... . .2nk k n nn n n nkS kC C C n C== = + + + 2 2 1 2 2 2 212 1 . .2 2 . .2 ... . .2nk k n nn n n nkS kC C C n C== = + + + =1 1( 1) 2 2n nk k k kn nk kkk C kC= = + Xt khai trin (1+x)n=0nk knkCx= ; n(1+x)n-1= 10nk knkkCx =Ly x=2ta cn.3n-1=102nk knkkC=2n.3n-1=02nk knkkC=

+n(n-1)(1+x)n-2 =20( 1)nk knkkk Cx = Ly x=2 ta c n(n-1)3n-2=20( 1) 2nk knkkk C=4n(n-1)3n-2=0( 1) 2nk knkkk C= Vy S=n.3n-2(2+4n) B. Theo chng trnh nng cao (2im) Cu VIa.1) Cho (P) y2 = x v ng thng (d): x y 2 = 0 ct (P) ti hai im A v B.Tm im C thuc cung AB sao choAABC c din tch ln nht +Ta A;B l nghim h: 22 0y xx y = = A(1;-1); B(4;2) +C(yo2;yo)e(P); h=d(C;d)=222o oy y +1 3.2 2ABCS h ABA= =22o oy y +Xt hm s f = 22o oy y Vi1 2oy s s Suy ra Max f = 9/4 Ti C(1/4;1/2) www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 252) Trong khng gian vi h ta Oxyz cho hai ng thng : 11 1( ) :2 1 1x y zd += = v

22 1( ) :1 1 1x y zd += =. Vit phng trnh mtphng cha (d1) v hp vi(d2) mt gc 300. Gi s mt phng cn tm l: 2 2 2( ) : 0 ( 0) ax by cz d a b c + + + = + + > . Trn ng thng (d1) ly 2 im: A(1; 0; -1), B(-1; 1; 0). Do( ) qua A, B nn: 0 20a c d c a ba b d d a b + = = + + = = nn( ) : (2 ) 0 ax by a bz a b + + + = . Yu cu bi ton cho ta: 02 2 2 2 2 21. 1. 1.(2 ) 1sin3021 ( 1) 1 . (2 )a b a ba b a b + = =+ + + + 2 2 2 22 3 2 3(5 4 2 ) 21 36 10 0 a b a ab b a ab b = + + =D thy0 b = nn chn b=1, suy ra: 18 1142118 11421aa

=

+=

KL: Vy c 2 mt phng tha mn: 18 114 15 2 114 3 114021 21 21x y z+ + + + =18 114 15 2 114 3 114021 21 21x y z ++ + = . Cu VII/b: Gii h phng trnh 20102 22( 1)log2 3xy xyy x x y = + + = 6 I. PHN CHUNG CHO TT C TH SINH (7 im). Cu I (2 im): Cho hm s 2 11xyx+=+ 1.Kho s t s bin thin v v th (C) ca hm s cho. 2.Tm tr n (C) nhng i m c tng khong c ch n haiti m cn ca (C) nh nht Cu II (2 im):1) Gii phng trnh: 2 22009cos 2 2 2 sin 4cos sin 4sin cos4x x x x x x | |+ + = + |\ .. 2) Gii h phng trnh: 2232 31 1(1 ) 414x xy yx xxy y y+ + + =+ + = . www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 26Cu III (1 im): Tnh tch phn: 0 22123 4 4. 2 14 4 5x xI x x dxx x| | = + +| |+ +\ .}. Cu IV (1 im):Trn ng thng vung gc ti A vi mt phng ca hnh vung ABCD cnh ata ly im S vi SA = 2a . Gi B, D l hnh chiu vung gc ca A ln SB v SD. Mt phng (ABD ) ct SC ti C . Tnh th tch khia dinABCDD C B. Cu V (1 im): Tam gi c ABC c c i m gnu c c gc tho m n: cos .cos cos .cos cos .cos 3cos cos cos 2+ + =A B B C C AC A B II. PHN RING CHO TNG CHNG TRNH ( 3 im). Th sinh ch c lm mt trong hai phn (phn 1 hoc phn 2) 1. Theo chng trnh Chun: 1.Cu VI.a (2 im): 1) Trong mt phng to Oxy , cho ng trn ( C) :2 22 6 15 0 x y x y + + =v ng thng (d) :3 0 mx y m =( m l tham s). Gi I ltm ca ng trn . Tm m ng thng (d) ct (C) ti 2 im phn bit A,B tho mn chu vi A IAB bng 5(2 2) +. 2) Trong khng gian vi h ta Oxyz cho hai ng thng : 11 1( ) :2 1 1x y zd += = v

22 1( ) :1 1 1x y zd += =. Vit phng trnh mtphng cha (d1) v hp vi(d2) mt gc 300. Cu VII.a (1 im):Chng minh rng vi a, b, c>0 ta c: 1 1 1 1 1 1 1 1 14 4 4 3 3 3 2 2 2 a b c a b b c c a a b c b c a c a b+ + > + + > + ++ + + + + + + + + 2. Theo chng trnh Nng cao: Cu VI.b (2 im) 1) Trong mt phng Oxy cho ng trn (C) tm I(-1; 1), bn knh R=1, M l mt im trn( ) : 2 0 d x y + = . Hai tip tuyn qua M to vi (d) mt gc 450 tip xc vi(C) ti A, B. Vit phng trnh ng thng AB. 2) Trong khng gian Oxyz cho t din ABCD bit A(0; 0; 2), B(-2; 2; 0), C(2; 0; 2), ( ) DH ABC v3 DH =vi H l trc tm tam gic ABC. Tnh gc gia (DAB) v (ABC). Cu VII.b (1 im):Chng minh rng vi a, b, c>0 ta c: 1( )( ) ( )( ) ( )( )a b ca a b a c b b a b c c c a c b+ + s+ + + + + + + + +. P N 6 I. PHN CHUNG CHO TT C TH SINH (7 im). Cu I (2 im): Cho hm s 2 11xyx+=+ 1.Kho st s bin thin v v th (C) ca hm s cho. 2.Tm tr n (C) nhng i m c tng khong c ch n haiti m cn ca (C) nh nht Cu II (2 im):1) Gii phng trnh: 2 22009cos 2 2 2 sin 4cos sin 4sin cos4x x x x x x | |+ + = + |\ .. www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 272 22009cos 2 2 2 sin 4cos sin 4sin cos4x x x x x x | |+ + = + |\ . 2 2cos sin 2(sin cos ) 4sin .cos (sin cos ) x x x x x x x x + + = +(cos sin )(cos sin 4cos .sin 2) 0 x x x x x x + + =cos sin 0 (1)cos sin 4sin .cos 2 0 (2)x xx x x x+ = + = + Gii (1):(1) tan 14x x k = = ++ Gii (2): tcos sin , 2 x x t t = sta c phng trnh: 22 0 t t + = . 01/ 2tt= = -Vi0 t =ta c:tan 14x x k = = +-Vi1/ 2 t = ta c: arccos( 2 / 4) / 4 2cos( ) 2 / 44arccos( 2 / 4) / 4 2x kxx k

= ++ = = + KL: Vy phng trnh c 4 h nghim: 4x k = + , 4x k = + , arccos( 2 / 4) / 4 2 x k = + ,arccos( 2 / 4) / 4 2 x k = + . 2).Gii h phng trnh: 2232 31 1(1 ) 414x xy yx xxy y y+ + + =+ + = . k 0 y =2 2223 33 2 31 1 1 1(1 ) 4 41 1 1( ) 4 4x x x xy y y yx x xx x xy y y y y y + + + = + + + = + + + = + + = t1a xyxby = + = Ta c 2 2 23 3 2 22 4 4 2 4 2 212 4 ( 4) 4 4 4 0a a b a a b a a b aba ab a a a a a a + = + = + = = = = + = + = Khi 111 2x yyx xx= = = + =KL Cu III (1 im): Tnh tch phn: 0 22123 4 4. 2 14 4 5x xI x x dxx x| | = + +| |+ +\ .}. 0 22123 4 4. 2 1(2 1) 4x xI x x dxx| | = + +| |+ +\ .}0 0 221 12 24 (2 1)( . 2 1)(2 1) 4xdx x x dxx += + ++ +} } www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 280 0 221 12 24 (2 1)( . 2 1)(2 1) 4xdx x x dxx += + ++ +} } + Tnh: 0 21 2124 (2 1)(2 1) 4xI dxx +=+ +}.t: 12 1 2sin , ; cos , 0, 02 2 2 6x t t dx tdt x t x t | |+ = e = = = = = |\ .. Khi : 2 2 6 6 6 61 2 2 20 0 0 02cos 2 1 sin 14sin 4 2(sin 1) 2 sin 1t tdt dtI dt dtt t t = = = ++ + +} } } } =62012 sin 1dtt ++} + Tnh: 6 62 2 20 0(tan )sin 1 2(tan 1/ 2)dt d tIt t = =+ +} }. t: 2tan tan2t y = . Suy ra: 22 2(tan ) (tan ) (1 tan )2 2d t d y y dy = = + , vi0 0,6t y t y = = = =sao cho 6tan3 =, (0 )2 < . Trn ng thng (d1) ly 2 im: A(1; 0; -1), B(-1; 1; 0). Do( ) qua A, B nn: 0 20a c d c a ba b d d a b + = = + + = = nn ( ) : (2 ) 0 ax by a bz a b + + + = . Yu cu bi ton cho ta: 02 2 2 2 2 21. 1. 1.(2 ) 1sin3021 ( 1) 1 . (2 )a b a ba b a b + = =+ + + + 2 2 2 22 3 2 3(5 4 2 ) 21 36 10 0 a b a ab b a ab b = + + =D thy0 b = nn chn b=1, suy ra: 18 1142118 11421aa

=

+=

KL: Vy c 2 mt phng tha mn: 18 114 15 2 114 3 114021 21 21x y z+ + + + =18 114 15 2 114 3 114021 21 21x y z ++ + = . Cu VII.a (1 im):Chng minh rng vi a, b, c>0 ta c: 1 1 1 1 1 1 1 1 14 4 4 3 3 3 2 2 2 a b c a b b c c a a b c b c a c a b+ + > + + > + ++ + + + + + + + + D c: 21 1 4( ) 4 ( , 0)(*) x y xy xyx y x y+ > + > + ++ + +. p dng 2 ln (*) ta c: 1 1 1 1 163 a b b b a b+ + + >+hay 1 3 163 a b a b+ >+(1) Tng t ta c: 1 3 163 b c b c+ >+(2) v 1 3 163 c a c a+ >+(3) Cng (1), (2) v (3) theo v vi v ri rt gn ta c iu phi chng minh. + Chng minh: 1 1 1 1 1 13 3 3 2 2 2 a b b c c a a b c b c a c a b+ + > + ++ + + + + + + + + p dng (*) ta c: 1 1 4 23 2 2( 2 ) 2 a b b c a a b c a b c+ > =+ + + + + + + (4) Tng t ta c: 1 1 2(5)3 2 2 b c c a b b c a+ >+ + + + +

1 1 2(6)3 2 2 c a a b c c a b+ >+ + + + + www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 31Cng (4), (5) v (6) theo v vi v ta c iu phi chng minh. 2. Theo chng trnh Nng cao: Cu VI.b (2 im) 1) Trong mt phng Oxy cho ng trn (C) tm I(-1; 1), bn knh R=1, M l mt im trn( ) : 2 0 d x y + = . Hai tip tuyn qua M to vi (d) mt gc 450 tip xc vi(C) ti A, B. Vit phng trnh ng thng AB. D thy( ) I d e . Hai tip tuyn hp vi (d) mt gc 450 suy ra tam gic MAB vung cn v tam gic IAM cng vung cn . Suy ra:2 IM = . ( ) ( M d M e a; a+2),( 1; 1) IM a a = + +, 02 2 1 22aIM aa= = + = = .Suy ra c 2 im tha mn: M1(0; 2) v M2 (-2; 0). + ng trn tm M1 bn kinh R1=1 l (C1): 2 24 3 0 x y y + + = . Khi AB i qua giao im ca (C ) v (C1) nn AB: 2 2 2 24 3 2 2 1 1 0 x y y x y x y x y + + = + + + + = . + ng trn tm M2 bn kinh R2=1 l (C2): 2 24 3 0 x y x + + + = . Khi AB i qua giao im ca (C ) v (C2) nn AB: 2 2 2 24 3 2 2 1 1 0 x y x x y x y x y + + + = + + + + + = . + KL: Vy c hai ng thng tha mn:1 0 x y + =v1 0 x y + + = . 2).Trong khng gian Oxyz cho t din ABCD bit A(0; 0; 2), B(-2; 2; 0), C(2; 0; 2), ( ) DH ABC v3 DH =vi H l trc tm tam gic ABC. Tnh gc gia (DAB) v (ABC). Trong tam gic ABC,giK CH AB = . Khi , d thy( ) AB DCK . Suy ra gc gia (DAB) v(ABC) chnh l gcDKH Z .Ta tm ta im Hri Tnh c HK l xong. + Phng trnh mt phng (ABC). -Vecto php tuyn( ) [ , ] 0; 4; 4 n ABAC = = -(ABC):2 0 y z + = . +( ) H ABC enn gi s( ; ; 2 ) Ha b b . Ta c:( ; ; ), (4; 2; 2). AH a b b BC = = ( 2; ; ), ( 2; 2; 2). CH a b b AB = = Khi : . 0 022 2 0. 0BC AH a ba ba bABCH= = = = + + == Vy H(-2; -2; 4). + Phng trnh mt phng qua H v vung gc vi AB l:4 0 x y z + = . Phng trnh ng thng AB l: 2x ty tz t= = = +. www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 32Gii h: 24 0x ty tz tx y z= = = + + =ta c x =2/3; y =-2/3, z =8/3. Suy ra: K(2/3;-2/3; 8/3). Suy ra: 2 2 22 2 8 962 2 43 3 3 3HK| | | | | |= + + + + = |||\ . \ . \ .. Gil gc cn tm th: tan / 96 / 12 6 / 3 arctan( 6 / 3) DH HK = = = =Vyarctan( 6 / 3) =l gc cn tm. Cu VII.b (1 im):Chng minh rng vi a, b, c>0 ta c: 1( )( ) ( )( ) ( )( )a b ca a b a c b b a b c c c a c b+ + s+ + + + + + + + +. V ia,b >0 ta c( )( )( )( ) ( )( )2 2 22a b a c ( ) 2 ( ) 0a b a c ( ) a b a c ( )( )( )ab ac a bc a bc a bcab ac ab aca a aa a b a c a ab ac a b c+ + + = + = > + + > + + + > + s =+ + + + + + + CM t2ri cng v vi v ta c dpcm 7 A.PHN CHUNG CHO TT C TH SINH(7.0 im)Cu I :( 2, 0 im) Cho hm s3 2y (m 2)x 3x mx 5 = + + + , m l tham s1.Kho st s bin thin v v th (C ) ca hm s khi m = 0 2.Tm cc gi tr ca m cc im cc i, cc tiu ca th hm s cho c honh l cc s dng.Cu II :( 2, 0 im) Gii cc phng trnh 1. 3 34sin x.c 3x 4cos x.sin3x 3 3c 4x 3 os os + + =2. 2 23 3 3log (x 5x 6) log (x 9x 20) 1 log 8+ + + + + =+Cu III :( 1, 0 im)Tm gi tr ca tch phn : 3e 31ln xIx 1 ln x=+} CuVI :( 1, 0 im) Mt mt phng qua nh S ca mt hnh nn ct ng trn y theo cung

AB c s o bngo. Mt phng (SAB) to vi y gc| . Bit khong cch t tm O ca y hnh nn n mt phng (SAB) bng a. Hy tm th tch hnh nn theoo, | v a CuV :( 1, 0 im). Cho x, y, z l ba s dng. Chng minh bt ng thc sau: 3 2 3 2 3 2 2 2 22 y 2 x 2 z 1 1 1x y y z z x x y z++ ++s+ + +

B.PHN RING (3,0 im) : Th sinh ch c lm mt trong hai phn(phn 1 hoc 2) 1.Theo chng trnh ChunCu VIa :(2,0 im)www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 331/ Trong mt phng (Oxy), cho ng trn (C ):2 22x 2y 7x 2 0 + = v hai im A(-2; 0), B(4; 3). Vit phng trnh cc tip tuyn ca (C ) ti cc giao im ca (C ) vi ng thng AB. 2/ Trong khng gian Oxyz, lp phng trnh mt phng (P) qua M(2; -1; 2) , song song vi Oy v vung gc vi mt phng (Q): 2x y + 3z + 4 = 0 Cu VIIa :(1,0 im) Cho cc ch s 1, 2, 3, 4, 5, 7, 9 . Hy cho bit c tt c bao nhius t nhin c 7 ch s khc nhau i mt sao cho hai ch s chn khng ng cnh nhau , c lp t cc ch s cho .2.Theo chng trnh Nng caoCu VIb :(2,0 im)1/ Trong mt phng (Oxy), cho ng trn (C ):2 22x 2y 7x 2 0 + = v hai im A(-2; 0), B(4; 3). Vit phng trnh cc tip tuyn ca (C ) ti cc giao im ca (C ) vi ng thng AB. 2/ Cho hm s 22x (m 1)x 3yx m+ + =+ . Tm cc gi tr ca m sao cho tim cn ca th hm stip xc vi parabol y = x2 +5 Cu VIIb :(1,0 im) Cho khai trin ( )x 13 x 12281log 3 1log 9 752 2 ++| |+ |\ .. Hy tm cc gi tr ca x bit rng s hng th6 trong khai trin ny l 224 P N 7 A.PHN CHUNG CHO TT C TH SINH(7.0 im)Cu I :( 2, 0 im) Cho hm s3 2y (m 2)x 3x mx 5 = + + + , m l tham s1.Kho st s bin thin v v th (C ) ca hm s khi m = 0 2.Tm cc gi tr ca m cc im cc i, cc tiu ca th hm s cho c honh l cc s dng.Cc im cc i, cc tiu ca th hm s cho c honh l cc s dng PT 2y' 3(m 2)x 6x m 0 == + + +c 2 nghim dng phn bit 2a (m 2) 0' 9 3m(m 2) 0' m 2m 3 0 3 m 1mm 0 m 0 3 m 2P 03(m 2)m 2 0 m 23S 0m 2= + = A = + > A = + > < < < < < < = > + + < < = > + Cu II :( 2, 0 im) Gii cc phng trnh 1. 3 34sin x.c 3x 4cos x.sin3x 3 3c 4x 3 os os + + =2 24 (1 cos x)sin x.cos3x (1 sin x)cos x.sin3x 3 3cos4x 3 [ ] + + = 4 sin x.cos3x cos x.sin3x) cos xsin x(cosx.cos3x sin x.sin3x) 3 3cos4x 3 [( ] + + + =1 14 sin 4x sin 2x. cos2x 3 3cos4x 3 4 sin 4x sin 4x 3 3cos4x 3 3sin 4x 3 3 cos4x 32 4[ ]| | + = + = + = |\ . 1 3 1sin 4x 3 cos4x 1 sin 4x cos 4x sin(4x ) sin2 2 2 3 6t t + = + = + = 4x k2 4x k2 4x k2 x k3 6 3 6 6 24 2(k Z)5 5x k 4x k2 4x k2 4x k28 2 3 6 3 6 2t t t t t t t+ = + t + = + t = + t = + et t t t t t t = + + = + t + = + t = + t

2. 2 23 3 3log (x 5x 6) log (x 9x 20) 1 log 8+ + + + + =+ (*) www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 34+ iu kin :22x 5x 5x 6 0 x 3 x 24 x 3x 5 x 4x 9x 20 0x 2< + + > < v >

< <

< v > + + > > , v c:3 31 log 8 log 24 + =+ PT (*) 2 2 2 23 3log (x 5x 6)(x 9x 20) log 24 (x 5x 6)(x 9x 20) 24(x 5) ( 4 x 3) (x 2)(x 5) ( 4 x 3) (x 2)( + + + + = + + + + = < v < < v > < v < < v > (x 2)(x 3)(x 4)(x 5) 24 (*)(x 5) ( 4 x 3) (x 2) (**) + + + + = < v < < v > + t 2t (x 3)(x 4) x 7x 12 (x 2)(x 5) t 2 = + + = + + + + = , PT (*) tr thnh : t(t-2) = 24 2(t 1) 25 t 6 t 4 = =v = -t = 6 : 2 2x 1x 7x 12 6 x 7x 6 0x 6= + + = + + = = ( tha kin (**)) -t = - 4 : 2 2x 7x 12 4 x 7x 16 0 + + = + + = : v nghim + Kt lun : PT c hai nghim l x = -1 v x = - 6 Cu III :( 1, 0 im)Tm gi tr ca tch phn : 3e 31ln xIx 1 ln x=+} + t 2dxt 1 ln x 1 ln x t 2tdtx= + + = =v ( )33 2ln x t 1 = + i cn : 3x 1 t 1 ; x e t 2 = = = =+ Tch phn 2 2 2 2 3 6 4 25 31 1 1(t 1) t 3t 3t 1 1I dt dt (t 3t 3t )dtt t t= + = = + } } } 6 4 221 3 3 15t t t ln t ln 21 6 4 2 4| |= + = |\ . CuVI :( 1, 0 im) Mt mt phng qua nh S ca mt hnh nn ct ng trn y theo cung

AB c s o bngo. Mt phng (SAB) to vi y gc| . Bit khong cch t tm O ca y hnh nn n mt phng (SAB) bng a. Hy tm th tch hnh nn theoo, | v a +Gi I l trung im ca dy cung AB v H l chn ng cao h t O ca tam gic SOI th : AB IO AB AB SI , AB SO (SIO) vAB OH ,vcIS OH theocch dng. T gi thit ca bi , ta c IOA OH a2 ; OIS; o= = | =+Cc tam gic SOI ( vung ti O) v IOH ( vung ti I) c SOH SIO = = | nn :

OH aOIsin sin= =| | v OH aOScos cos= =| | +Tam gic OIA vung ti I v

IOA2o=nn b/knh ng trn y l OI acos sin .cos2 2R OA = = =o o| + Th tch hnh nn l : 2322 21 1 a a aV ). .3 3 co ssin .co s 3sin .co s .co s2 2(R OS| | || | t= t = t = ||o o|\ . | | | |\ . CuV :( 1, 0 im). Cho x, y, z l ba s dng. Chng minh bt ng thc sau: www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 35 3 2 3 2 3 2 2 2 22 y 2 x 2 z 1 1 1x y y z z x x y z++ ++s+ + +

CM bt ng thc 3 2 3 2 3 2 2 2 22 y 2 x 2 z 1 1 1x y y z z x x y z++ ++s+ + +vi x > 0 ; y > 0 ; z > 0 + p dng bt ng thc Csi cho hai s dng x3 v y2ta c : 3 2 3 23 22 x 1x y 2 x y 2 xxyx y xy+ > = s+, du ng thc xy ra khi v ch khi x3 = y2(1) Tng t :3 22 y 1y z yzs+, du ng thc xy ra khi v ch khi y3 = z2(2)

3 22 z 1z x zxs+, du ng thc xy ra khi v ch khi z3 = x2(3) + p dng BT(d CM ) 2 2 2ab bc ca a b c + + s + + (du ng thc xy ra khi v ch khia = b = c )ta c : 2 2 21 1 1 1 1 1xy yz zx x y z++ ++s, du ng thc xy ra khi v ch khi x = y = z(4) + T (1), (2), (3) v (4) ta c BT cn C/minh . Du ng thc xy ra khi v ch khix = y = z > 0 B.PHN RING (3,0 im) : Th sinh ch c lm mt trong hai phn(phn 1 hoc 2) 1.Theo chng trnh ChunCu VIa :(2,0 im)1/ Trong mt phng (Oxy), cho ng trn (C ):2 22x 2y 7x 2 0 + = v hai im A(-2; 0), B(4; 3). Vit phng trnh cc tip tuyn ca (C ) ti cc giao im ca (C ) vi ng thng AB. + ng trn (C ) :22 2 2 2 27 7 652x 2y 7x 2 0 x y x 1 0 x y2 4 16| |+ = + = + = |\ . (C ) c tm7I ; 04| | |\ . v bn knh 65R4=+ ng thng AB vi A(-2; 0) v B(4; 3) c phng trnhx 2 y x 2y6 3 2, hay : + += =+ Giao im ca (C ) vi ng thng AB c ta l nghim h PT22 2 2x 25x(x 2) 0 2x 2y 7x 2 0 2x 2 7x 2 0x 0; y 12x 2 x 2x 2; y 2x 22 22y = y =y =+ | | = + = + = |= = \ . + + = =+ Vy c hai giao im l M(0; 1) v N(2; 2) + Cc tip tuyn ca (C ) ti M v N ln lt nhn cc vect 7IM ;14| |= |\ . v 1IN ; 24| |= |\ . lm cc vect php tuyn , do cc TT c phng trnh ln lt l : - 7(x 0) 1(y 1) 0 7x 4y 4 04 , hay : + = + =- 1(x 2) 2(y 2) 0 x 8y 18 04 , hay : + = + =2/ Trong khng gian Oxyz, lp phng trnh mt phng (P) qua M(2; -1; 2) , song song vi Oy v vung gc vi mt phng (Q): 2x y + 3z + 4 = 0 www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 36Cch 1 + Mt phng (Q) : 2x y + 3z + 4 = 0 c VTPT( )Qn 2; 1;3 = v trc Oy c VTV ( ) j 0 ; 1 ; 0 =. Hai vect Qnvj khng cng phng vi nhau. + Gi Pnl VTPT ca mt phng (P) . V (P) song song vi Oy v vung gc vi mt phng (Q) nn QPn n v Pn j , do c th chn P Qn j, n (3; 0; 2) (= = .Mp i qua M v c VTPT (3; 0; 2) l 3(x - 2) + 0(y+1) -2(z - 2) = 0 , hay l :3x - 2z- 2 = 0 // Oy. Vy (P) : 3x - 2z- 2 = 0Cch 2 + Mt phng (P) song song trc Oy v i qua M( 2; -1; 2) nn c phng trnh dng :a( x 2 ) + c(z 2) = 0ax cz 2a 2c 0 + =, vi 2 2a c 0 + =v2a 2c 0 =+ Mt phng (P) vung gc vi mt phng (Q) : 2x y + 3z + 4 = 0 nn c 2a + 3c = 0 : chn a = 3 v c = -2 , khi -2a 2c =2 0 = , do PT mp(P) l : 3x 2z 2 = 0Cu VIIa :(1,0 im) Cho cc ch s 1, 2, 3, 4, 5, 7, 9 . Hy cho bit c tt c bao nhius t nhin c 7 ch s khc nhau i mt sao cho hai ch s chn khng ng cnh nhau , c lp t cc ch s cho . t A = { 1, 2, 3, 4, 5, 7, 9 }+ Tng s cc s t nhin c 7 ch s khc nhau i mt lp c t cc ch s ca tp A l 7! + Trong A c hai ch s chn l 2 v 4 nn : Tng s cc s t nhin c 7 ch s khc nhau i mt sao cho hai ch s chn lun ng cnh nhau , lp c t cc ch s ca tp A l : 2!6! + Vy : Tng cc s t nhin tha mn yu cu bi ton l : 7! 2!6! = 6!(7 2) = 6!5 = 3600 (s )2.Theo chng trnh Nng caoCu VIb :(2,0 im)1/ Trong mt phng (Oxy), cho ng trn (C ):2 22x 2y 7x 2 0 + = v hai im A(-2; 0),B(4;3).Vitphngtrnhcctiptuynca(C)ticcgiaoimca(C)vingthng AB. 2/ Cho hm s 22x (m 1)x 3yx m+ + =+ . Tm cc gi tr ca m sao cho tim cn ca th hm stipxc vi parabol y = x2 +5 Hm s 22x (m 1)x 3yx m+ + =+ xc nh vi mix m = Vit hm s v dng 2m m 3y 2x 1 mx m = + ++ + TH1 : 21 13m m 3 0 m2 = =: C hm s bc nhty 2x 1 m = + ( x m = ) : th khng c tim cn+ TH2 : 21 13m m 3 0 m2 = =: th hm s c tim cn ng l ng thng (d1) x = -mv tim cn xin l ng thng (d2)y = 2x + 1 - m+ ng thng (d1) x = - m lun ct parabol parabol y = x2 +5 ti im (-m ; m2 +5) ( vi mi 1 13m2= ) v khng th l tip tuyn ca parabol www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 37+ Tim cn xin (d2) y = 2x + 1 - m tip xc vi parabol y = x2 +5PT x2 +5= 2x + 1 - m , hay PT x2 2x + 4 +m= 0 c nghim kp ' A =1-(4 + m) = 0m 3 = ( tha iu kin) Kt lun : m = -3 l gi tr cn tm Cu VIIb :(1,0 im) Cho khai trin ( )x 13 x 12281log 3 1log 9 752 2 ++| |+ |\ .. Hy tm cc gi tr ca x bit rng s hng th6 trong khai trin ny l 224Ta c :( )k 88k 8 k k8k 0a b C a b==+ = vi ( )( )( )x 13 x 1221 1 1log 3 1log 9 7 x 1 x 15 3 5a 2 9 7 b 2 3 1 =; + + = + = = ++ Theo th t trong khai trin trn , s hng th su tnh theo chiu t tri sang phi ca khai trin l

( ) ( ) ( ) ( )3 51 115 x 1 x 1 x 1 x 13 56 8T C 9 7 . 3 1 56 9 7 . 3 1 | || |= + + = + + | |\ .\ . + Theo gi thit ta c : ( ) ( )x 11x 1 x 1 x 1 x 1x 19 756 9 7 . 3 1 4 9 7 4(3 1)3 1= 224 ++ + = + = ++ ( )x 12x 1 x 1x 13 1 x 13 4(3 ) 3 0x 23 3 = = + =

== 8 I-PHN CHUNG CHO TT C CC TH SINH :(7 im) Bi I.. (2 im) Cho hm s 3 21 23 3y x mx x m = + +c th (Cm) a)Kho st khi m =-1. b)Tm m (Cm) ct Ox ti 3 im phn bit c tng bnh phng cc honh ln hn 15. Bi II .(2 im). Cho phng trnh 3 3cos si n x x m =(1) a)Gii phng trnh khi m=-1 b)Tm m phng trnh (1) c ng hai nghim;4 4x (e ( Bi III. (2 im) a)Gii phng trnh 2 2 2l og 9 l og l og 3 2.3xx x x = b)Tnh tch phn 2 44 24si ncos (tan 2tan 5)xdxx x x +} Bi IV(2 im) a/.Cho khai trin ( )52 3 150 1 151 ... x x x a a x a x + + + = + + + . Tm h s 9aca khai trin . b/.Cho a, b, c>0; abc=1. Chng minh rng 3 3 33(1 )(1 ) (1 )(1 ) (1 )(1 ) 4a b cb c c a a b+ + >+ + + + + +. II-PHN RING(3im)( Th sinh ch lm cu Va hoc Vb) Bi Va.(3 im). www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 381. Trong khng gian Oxyz, cho mt cu (S) c phng trnh ( ) ( ) ( )2 2 21 2 3 14 x y z + + + + + =v im ( )1; 3; 2 M . Lp phng trnh mt phng (P) i qua sao cho (P) ct (S) theo mt giao tuyn l ng trn c bn knh nh nht. 2.Trong mt phng Oxy, cho im ( )1; 3 Anm ngoi (C): 2 26 2 6 0 x y x y + + + = . Vit phng trnh ng thng d qua A ct (C) ti hai im B v C sao cho AB=BC Bi Vb.(3 im). 1. (1 im) Trong mt phng vi h to Oxycho tam gicABCvung A. Bit ( ) ( ) 1; 4 , 1; 4 A B v ng thngBCi qua im 12;2M | | |\ .. Hy tm to nhC . 2.(2im)ChohnhchpS.ABCDcyABCDlhnhvungc nha,SAvunggcv imt phng (ABCD) v SA=2a. Gi E l trung im ca cnh CD.a)Tnh theo a khong cch t im S n ng thng BE. b)Tm tm v bn knh mt cu ngoi tip hnh chp S.ABCD. P N 8 I-PHN CHUNG CHO TT C CC TH SINH :(7 im) Bi I.. (2 im) Cho hm s 3 21 23 3y x mx x m = + +c th (Cm) a)Kho st khi m =-1. b)Tm m (Cm) ct Ox ti 3 im phn bit c tng bnh phng cc honh ln hn 15. YCBT tha 3 21 203 3x mx x m + + =c 3 nghim phn bit tha 2 2 21 2 315 x x x + + > . ( )( )21 (1 3 ) 2 3 0 x x mx m + + + =c 3 nghim phn bit tha 2 2 21 2 315 x x x + + > . 1 m >. Bi II .(2 im). Cho phng trnh 3 3cos si n x x m =(1) a)Gii phng trnh khi m=-1 Khi m=-1, phng trnh tr thnh ( )( )cos si n 1 cos si n 1 x x x x + = t t =cos si n x x ; iu kin2 t s . Ta c nghim ( )2,22x kklx l

= +

e

= +

b)Tm m phng trnh (1) c ng hai nghim;4 4x (e ( (1) ( )( )cos si n 1 cos si n x x x x m + = t t =cos si n x x ; iu kin2 t s .Khi; 0; 24 4x t ( (e e ( . Ta c phng trnh theo t: 33 2 t t m = . www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 39Bng cch tm tp gi tr hm v tri, ta suy ra phng trnh c ng hai nghim ;4 4x (e ( khi v ch khi 2;12m |e | |

.. Bi III. (2 im) a)Gii phng trnh 2 2 2l og 9 l og l og 3 2.3xx x x = K: x>0.Ta c phng trnh 2 2 2 2l og 9 l og l og 3 l og 2 2.3 3 1x xx x x x = = . t 2l og 2tx x = .Phng trnh tr thnh 3 13 4 1 1 1 24 4t tt tt x| | | |= + = = = ||\ . \ . b)Tnh tch phn 2 44 24si ncos (tan 2tan 5)xdxx x x +} 2 44 24si ncos (tan 2tan 5)xdxIx x x= +}. t 2tan1dtt x dxt= =+. Ta c 1 1 22 21 122 l n 33 2 5 2 5t dt dtIt t t t = = + + +} } Tnh 11 212 5dtIt t= +}. t 0141 1tan2 2 8tu I du= = =}.Vy 2 32 l n3 8I= + . Bi IV(2 im) a)Cho khai trin ( )52 3 150 1 151 ... x x x a a x a x + + + = + + + . Tm h s 9aca khai trin . ( ) ( )( )5 55 52 3 25 100 01 1 1k m k mk mx x x x x CC x += = (+ + + = + + =

do 9acho tng ng k+m=9. Suy ra 0 9 1 8 2 7 3 6 4 5 5 45 10 5 10 5 10 5 10 5 10 5 109 5005 a CC C C CC CC CC CC = + + + + + =. b/.Cho a, b, c>0; abc=1. Chng minh rng 3 3 33(1 )(1 ) (1 )(1 ) (1 )(1 ) 4a b cb c c a a b+ + >+ + + + + +. p dng bt ng thc csi cho ba s, ta c + ++ + >+ +31 1 3(1 )(1 ) 8 8 4a c b ab c + ++ + >+ +31 1 3(1 )(1 ) 8 8 4b c a bc a ( )+ ++ + >+ + + > + +31 1 3(1 )(1 ) 8 8 43 1(1)4 2c a b ca bVT a b c www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 40Du bng xy ra khi 1 1 118 8 81a c ba b cabc + + += = = = ==. Vy 3 3 3(1) (1)2 4 4VT VT > > pcm. II-PHN RING(3im)( Th sinh ch lm cu Va hoc Vb) Bi Va.(3 im). 1. Trong khng gian Oxyz, cho mt cu (S) c phng trnh ( ) ( ) ( )2 2 21 2 3 14 x y z + + + + + =v im ( )1; 3; 2 M . Lp phng trnh mt phng (P) i qua sao cho (P) ct (S) theo mt giao tuyn l ng trn c bn knh nh nht. Ta thy M thuc min trong ca (S) v (S) c tm ( )1; 2; 3 , 14 I R = . Do ,(P) qua M ct (S) theo mt giao tuyn l ng trn c bn knh nh nht 2 2R I H nh nht (H l hnh chiu vung gc ca I trn mt phng (P)) I H ln nht ( )0;1; 1 M H I M = l VTPT ca (P). Vy (P) c phng trnh l y-z+1=0. 2.Trong mt phng Oxy, cho im ( )1; 3 Anm ngoi (C): 2 26 2 6 0 x y x y + + + = . Vit phng trnh ng thng d qua A ct (C) ti hai im B v C sao cho AB=BC Theo yu cu bi ton, , AB C thng hng v AB=BC.Gi 2 1( ; ), ( ; )2 1m aBab Cmnn b = = . Do B, C nm trn (C) nn 2 22 236 2 6 0 15 6 2 6 01aa b a b bm m n m nn =+ + + = = = + + + = = hoc 751595135abmn = == = . Vy c hai ng thng tha mn yu cu bi ton l x+y-4=0 v 7x+y-10=0. Bi Vb.(3 im). 1. (1 im) Trong mt phng vi h to Oxycho tam gicABCvung A. Bit ( ) ( ) 1; 4 , 1; 4 A B v ng thngBCi qua im 12;2M | | |\ .. Hy tm to nhC . tBCi qua( ) 1; 4 B v 12;2M | | |\ . nn c pt: 1 4912x y += 9 2 17 0 x y =9 17; ,2tC BC C t t | |e e |\ . ( )9 252; 8 ; 1;2tAB AC t | |= = + |\ . . V tam gicABCvung tiAnn. 0 AB AC = Suy ra 9 251 4. 0 3.2tt t+ = =Vy( ) 3; 5 C 2.(2 im)Cho hnh chp S.ABCD c y ABCD l hnh vung cnh OIFEABDCSwww.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 41IFEADBCa, SA vung gc vi mt phng (ABCD) v SA=2a. Gi E l trung im ca cnh CD.a)Tnh theo a khong c ch t im S n ng thng BE. Gi F l trung im ca BC => AFBE VABF AIB A A 2 2 2222 25 54AI AB AB a a aAIAB AF AF aaa = = = = =+ VAI BESI BESA BE `)

22 2 24 2 645 5a aSI SA AI a = + = + =b)Tm tm v bn knh mt cu ngoi tip hnh chp S.ABCD. Gi O l trung im ca SC (1) SO AO CO = = VSDC A vung gc D ( , CD SD CD AD )SO OD =(2) VSBC A vung gc B ( , BC BA BC SA ) SO OB =(3) T (1), (2), (3) SO=AO=BO=CO=DO => O l tm ca mt cu ngoi tip hnh chp =>2 2 2 22 4 62 2 2 2SO AC SA a a aR+ += = = = 9 Cu 1. Cho hm s 22xyx=+ 1.Kho st s bin thin v v th (C) ca hm s cho. 2.Vit phng trnh tip tuyn ca th (C), bit rng khong cch t tm i xng ca th (C) n tip tuyn l ln nht. Cu 2: ( 2 i m) Cho h phng trnh: 2 221x xy y mxy xy m+ + = + + = + 1) Gii h phng trnh vi3 m = . 2) Tmm h phng trnh c nghim duy nht. Cu 3:1). Cho a, b, c l cc s thc dng tho mn3 = + + c b a .Chng minh rng: 13 4 ) ( 32 2 2> + + + abc c b a2).ChoABC A . Chng mi nh rng:sin 2 sin 2 sin 2 4sin sin sin A B C A B C + + =Trong , , A B Cl ln ba gc caABC Aidi n ln lt vi ba cnh, , BC CAAB . Cu 4: ( 2i m) www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 42ChoABC Avung gc t i A. Trn ng thng vung gc vi mt phng( ) ABCt i Bta l y mt i mSsao cho1 SB BA AC = = = .( ) Pl mt phng song song vi cc cnhSBvACct c c c nh, , , SA SCBCBA ln lt ti, , , DEFH . 1) Chng mi nh rng:DEFHl hnh ch nht. 2) X c nh vtrca mt phng( ) Psao cho di n tch hnh ch nht l n nht. Cu 5: ( 1i m) 1).Cho, , a b cl cc s thc khng m, imt kh c nhau. Chng mi nh rng: 2ab bc ca a b ca b b c c a+++ + s+ + + Hi du = xy ra khi no? 2). Cho x, y, z l ba s thc dng thay i v tha mn:xyz z y x s + +2 2 2. Hy tm gi tr ln nht ca biu thc: xy zzzx yyyz xxP+++++=2 2 2. 3).Gii h phng trnh sau: =++=++ + +3127) (3) ( 4 422 2y xxy xy x xy P N 9 Cu I. Cho hm s 22xyx=+ 1).Kho st s bin thin v v th (C) ca hm s cho. 2).Vit phng trnh tip tuyn ca th (C), bit rng khong cch t tm i xng ca th (C) n tip tuyn l ln nht. Tip tuyn ca th (C) ti im M c honh 2 a = thuc th (C) c phng trnh: ( ) ( ) ( ) ()2224 24 2 2 022ay x a x a y a daa= + + + =++ Tm i xng( ) 2; 2 I . Ta c ( )( ) ( )4 28 2 8 2 8 2, 2 22 2 216 2 2.4. 2a a ad I daa a+ + += s = =++ + + ( ) , d I dln nht khi( )202 44aaa= + = = T suy ra c hai tip tuyn y = x v y = x + 8 Cu 2: ( 2 i m) Cho h phng trnh: www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 432 221x xy y mxy xy m+ + = + + = + 1) Gii h phng trnh vi3 m = . Nhn thy rng y l h phng trnh i xng loi I. Khi : t,x y Sxy P+ = = i u ki n 24 0 S P >Vit li h phng trnh di dng: ( )( )221 1x y xy mS P mSP m x y xy m+ + = + + = + = + + = + ( ) IKhi, SP l nghim ca phng trnh bc hai: ( )212 1 01tt m t mt m= + + + = = +

1111x yxy mx y mxy + = = +

+ = +

=

( ) ( )( ) ( ) ( )221 11 1 2f u u u mg u u m u = + + = + +

V im=-3, ta c: ( )21 1; 21 2 02 2; 1u x yu uu x y= = = = = = = ( )22 1 0 1 1 u u u x y + + = = = = Vy: v i 3 m = , h phng trnh cho c ba cp nghim l:( ) ( ) ( ) 1; 2 , 2; 1 , 1; 1 . 2) Tmm h phng trnh c nghim duy nht. i u ki n cn:Nhn xt rng nu hc nghi m( )0 0; x y th( )0 0; y xcng l nghim ca h, do hc nghi m duy nht khi0 0x y = . Khi: 2 30 0 03 3 20 0 0 012 2 2 132 1 2 2 1 034mx x m m xmx m x x xm= + = + = = = + + = = i u ki n :+V i 1 m = , ta c: www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 44( )( )( )32xy x yIxy x y+ + = + = Khi, xyl nghim ca phng trnh:

( )212 13 2 0 12 21x yvnxy tt t x yt x yxy + = = =

+ = = =

= + =

=

l nghim duy nht ca h. +V i 1 m = , ta c: ( )( )( )10xy x yIxy x y+ + = + =. Nhn thy hl un c cp nghi m( ) 0;1v( ) 1; 0 . +V i34m = , ta c: ( )( )( )5414xy x yIxy x y+ + = + = Khix y +vxyl nghim ca phng trnh: ( )21115 1 1 4014 4 2 1441x yxy tt t x ytx yvnxy + =

= =

+ = = =

=

+ = = l nghim duy nht ca h. Vy: v i 1 m =hoc 34m = hcho c nghi m duy nht. Cu 3: (1i m) 1). Cho a, b, c l cc s thc dng tho mn3 = + + c b a .Chng minh rng: 13 4 ) ( 32 2 2> + + + abc c b at 2; 13 4 ) ( 3 ) , , (2 2 2c bt abc c b a c b a f+= + + + =*Trc ht ta chng minh:) , , ( ) , , ( t t a f c b a f > :Tht vyDo vai tr ca a,b,c nh nhau nn ta c th gi thitc b a s s 3 3 = + + s c b a ahay a 1 s = ) , , ( ) , , ( t t a f c b a f 13 4 ) ( 3 13 4 ) ( 32 2 2 2 2 2 2+ + + + + + at t t a abc c b awww.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 45=) ( 4 ) 2 ( 32 2 2 2t bc a t c b + +=((

+ +((

+ +2 22 24) (44) ( 23c bbc ac bc b =22) (2) ( 3c b ac b = 02) )( 2 3 (2> c b a do a 1 s *By gi ta ch cn chng minh:0 ) , , ( > t t a fvia+2t=3 Ta c13 4 ) ( 3 ) , , (2 2 2 2 + + + = at t t a t t a f = 13 ) 2 3 ( 4 ) ) 2 3 (( 32 2 2 2 + + + t t t t t = 0 ) 4 7 ( ) 1 ( 22> t tdo 2t=b+c < 3 Du = xy ra1 0 & 1 = = = = = c b a c b t (PCM) 2).ChoABC A . Chng mi nh rng: sin 2 sin 2 sin 2 4sin sin sin A B C A B C + + =Trong , , A B Cl ln ba gc caABC Ai din ln lt vi ba cnh, , BC CAAB . TrongABC A , ta c: 2 2B C AA B C B C A + + + = + = =Khi: ( ) ( )( )( )sin 2 sin 2 sin 2 2sin cos 2sin cos2sin cos 2sin cos2sin cos cos2sin cos cos2 22sin cos cos2 22sin sin sinA B C A A B C B CA A A B CA A B CA B C A B CAA C BA B C dpcm + + = + + = + = + ( + +=| | | |= ||\ . \ .= Cu 4: ( 2i m) ChoABC Avung gc t i A. Trn ng thng vung gc vi mt phng( ) ABCt i Bta l y mt i mSsao cho1 SB BA AC = = = .( ) Pl mt phng song song v ic c c nhSBvACct c c c nh , , , SA SCBCBA ln lt ti, , , DEFH . 1).Chng mi nh rng:DEFHl hnh ch nht. ( ) ( )( )/ // /P SAB DHDH SBP SB = (1) ABCSD EFHwww.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 46( ) ( )( )/ // /P SBC EFEF SBP SB = (2) ( ) ( )( )/ // /P SAC DEDE ACP AC = (3) ( ) ( )( )/ // /P ABC HFHF ACP AC = (4) T (1), (2), (3), (4) suy ra t gi cDEFHl hnh bnh hnh(5) Mt kh c: ( )/ /SB ABC SB HFDH HFSB DH (6) Vy: T (5) v (6) suy raDEFHl hnh ch nht. 2). X c nh vtrca mt phng( ) Psao cho di n tch hnh ch nht l n nht Hnh ch nhtDEFHsc di n tch l n nht( ) P iqua bn i m, , , DEFHln lt l trung i m ca c c c nh, , , SA SCBCAB. Cu 5: ( 1i m) Cho, , a b cl cc s thc khng m, i mt khc nhau. Chng minh rng: 2ab bc ca a b ca b b c c a+++ + s+ + + Hi du = xy ra khi no? p dng bt ng thc Csicho c c s, , a b cta c: ( )( )( )2 12 22 22 22 32 2ab ab ab aba b aba b a b abbc bc bc bcb c bcb c b c bcca ca ca cac a cac a c a ca+ > s s+ ++ > s s+ ++ > s s+ + Cng vv ivca( ) ( ) ( ) 1 , 2 , 3ta c: 212 2 2 22ab bc ca ab bc caa b b c c aab bc ca a b b c c aa b b c c aab bc ca a b cdpcma b b c c a+ ++ + s+ + ++ + + | | + + s + + |+ + +\ .++ + + s + + + Du = xy ra khi v ch khia b c = = . www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 472). Cho x, y, z l ba s thc dng thay i v tha mn:xyz z y x s + +2 2 2. Hy tm gi tr ln nht ca biu thc: xy zzzx yyyz xxP+++++=2 2 2. V0 ; ; > z y x , p dng BT Csi ta c: xy zzzx yyyz xxP2 2 22 2 2+ + s = ||.|

\|+ + =xy zx yz2 2 241

||.|

\| + +s||.|

\| + +=||.|

\|+ + + + + sxyzz y xxyzxy zx yzy x x z z y2 2 22121 1 1 1 1 1 1412121=||.|

\|sxyzxyz Du bng xy ra3 = = = z y x . Vy MaxP = 21 3).Gii h phng trnh sau: =++=++ + +3127) (3) ( 4 422 2y xxy xy x xy K: x + y=0 Ta c h 2 2233( ) ( ) 7( )13x y x yx yx y x yx y+ + + =+ + + + =+ t u = x + y + 1x y + (2 u > ) ; v = x y ta c h : 2 23 133u vu v + = + = Gii h ta c u = 2, v = 1 do (2 u > ) T gii h 12 1 11 01x y x y xx yx y yx y + + = + = = + = = = 10 Phn c hung c ho t tc t hsi nh (7,0 i m) Cu I(2 i m) Cho hm s 2 11xyx+=+ 1.Kho st s bin thin v v th (C) ca hm s cho. 2.Tm tr n (C) nhng i m c tng khong c ch n haiti m cn ca (C) nh nht. Cu I I(2 i m) 1.Gii h phng trnh: 1 1 46 4 6x yx y + + =+ + + = 2.Gii phng trnh: 1 2(cos sin )tan cot 2 cot 1x xx x x=+ www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 48Cu I I I(1 i m) Trong mt phng (P) cho ng trn (C) tm O ng knh AB = 2R.Trn ng thng vung gc v i(P) t iO l y i m S sao cho OS = R 3 . I l im thuc on OS vi SI = 23R. M l mt im thuc (C). H l hnh chiu ca I trn SM. Tm v tr ca M tr n (C) t di n ABHM c thtch l n nht.Tm gi trl n nht .Cu I V (1 i m)Tnh tch phn: I= 121 1 1dxx x + + +} Cu V (1 i m) Cho x, y, z l 3 s thc dng tha mn xyz=1. Chng minh rng

1 1 111 1 1 x y y z z x+ + s+ + + + + + Phn r i ng (3,0 i m).Th sinh ch c lm mt trong hai phn (phn A hoc B) A.Theo chng trnh Chun Cu VI .a (1 i m) Trong mt phng Oxy cho tam gi c ABC bi t A(2; - 3), B(3; - 2), c di n tch bng 32 v trng tm thuc ng thngA: 3x y 8 = 0. Tm ta nh C. Cu VI I .a (1 i m) T cc ch s 0,1,2,3,6,7,8,9 c th lp c bao nhiu s t nhin c 6 ch s imt kh c nhau ( ch s u ti n phikh c 0) trong phic ch s 7.Cu VI I I .a (1 i m) Tm a bt phng trnh sau c nghim:21 13 3log 1 log ( ) x ax a + > + B.Theo chng trnh Nng cao Cu VI .b (1 i m) Trong mt phng Oxy cho el i p (E): 2 214 3x y+ = v ng thngA:3x + 4y =12. T i m M bt ktr nA k ti (E) cc tip tuyn MA, MB. Chng minh rng ng thng AB l un iqua mt i m c nh. Cu VI I .b (1 i m)Cho hm s 24 32x xyx+ +=+ c th (C).Gi s ng thng y = kx + 1 ct (C) t i2 i m phn bi t A, B. Tm tp h p trung i m Ica AB khik thay i . Cu VI I I .b (1 i m)1).Gii phng trnh: ( ) ( )2 2 2log log3 1 . 3 1 1x xx x + + = +2). Cho cc s thc a, b, c tha mn :0 1, 0 1, 0 1 a b c < s < s < s . Chng minh rng:( )1 1 1 11 3 a b cabc a b c| |+ + + >+ + + |\ . 3). Gii h phng trnh: 3 1 2 322 2 3.23 1 1x y y xx xy x+ + + =+ + = + P N 10 Phn c hung c ho t tc t hsi nh (7,0 i m) Cu I(2 i m) Cho hm s 2 11xyx+=+ 1.Kho st s bin thin v v th (C) ca hm s cho. 2.Tm tr n (C) nhng i m c tng khong c ch n haiti m cn ca (C) nh nht. www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 49GiM(x0;y0) l mt im thuc (C), (x0= - 1) th0002 11xyx+=+ GiA, B l n l t l hnh chiu ca M trn TC v TCN th MA = |x0+1| , MB = | y0- 2| = |002 11xx++- 2| = |011 x +| Theo Cauchy thMA + MB> 2001x 1.1 x++=2 MA + MB nh nht bng 2 khix0 = 0 hoc x0 = -2.Nh vy ta c hai im cn tm l (0;1)v (-2;3) Cu I I(2 i m) 1.Gii h phng trnh: 1 1 46 4 6x yx y + + =+ + + = i u ki n: x >-1, y >1 Cng vtheo vritr vtheo vta c h1 6 1 4 106 1 4 1 2x x y yx x y y + + + + + + =+ + + + = t u= 1 6 x x + + + , v = 1 4 y y + + . Ta c h105 52u vu v+ =+ = {55uv== {35xy== l nghim ca h 2. Gii phng trnh: 1 2(cos sin )tan cot 2 cot 1x xx x x=+ iu kin:sinx.cosx = 0 vcotx= 1 Phng trnh tng ng 1 2(cos sin )sin cos 2 cos1cos sin 2 sinx xx x xx x x=+ cosx = 22 x = 24k +i chiu iu kin pt c 1 h nghim x = 24k +Cu I I I(1 i m)Trong mt phng (P) cho ng trn (C) tm O ng knh AB = 2R.Trn ng thng vung gc vi (P) t iO l y i m S sao cho OS = R 3 . I l i m thuc o n OS v iSI= 23R. M l mt im thuc (C). H l hnh chi u ca Itr n SM. Tm vtrca M tr n (C) t di n ABHM c thtch l n nht.Tm gi trl n nht .SHIOBMAwww.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 50T gic IHMO ni tip nn SH.SM = SI.SO m OS = R 3 ,SI= 23R, SM = 2 22 SO OM R + = SH = R hay H l trung imca SM Gi K l hnh chiu vung gc ca H ln mp(MAB) thHK = 12SO=32R , (khng i ) VBAHM l n nht khidt( A MAB) l n nhtM l imgia ca cung AB Khi VBAHM=336R (vtt) Cu I V (1 i m)Tnh tch phn: I= 121 1 1dxx x + + +} t u = x+21 x + thu - x= 21 x + 2 2 22 1 x ux u x + = +221 1 112 2ux dx duu u | |= = + |\ . icn x= - 1 thu = 2 -1 x = 1 thu =2 +1 2 1 2 1 2 1222 1 2 1 2 11 111 1 21 2 1 2 (1 )dudu du uIu u u u+ + + | |+ |\ .= = ++ + +} } } =2 1 2 122 1 2 11 1 1 1 112 1 2 1duduu u u u+ + | |+ + = |+ +\ .} } Cu V (1 i m) Cho x, y, z l 3 s thc dng tha mn xyz=1. Chng minh rng

1 1 111 1 1 x y y z z x+ + s+ + + + + + t x=a3y=b3 z=c3 th x, y, z >0 v abc=1.Ta c a3 + b3=(a+b)(a2+b2-ab) >(a+b)ab, do a+b>0 v a2+b2-ab >ab a3 + b3+1 > (a+b)ab+abc=ab(a+b+c)>0 ( )3 31 1ab 1 ab a b cs+ + + + Tng t ta c ( )3 31 1 c 1 bc a b c bs+ + + +, ( )3 31 1 a 1 ca a b c cs+ + + + Cng theo vta c www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 511 1 11 1 1 x y y z z x+ ++ + ++ + +=3 31ab 1 + ++3 31 c 1 b + ++3 31 a 1 c + + s ( )1 1 1 1a b c ab bc ca| |+ + |+ +\ .=( )( )11a b cc a b + + =+ + Du bng xy ra khix=y=z=1 Phn r i ng (3,0 i m).Th sinh ch c lm mt trong hai phn (phn A hoc B) A.Theo chng trnh Chun Cu VI .a (1 i m) Trong mt phng Oxy cho tam gi c ABC bi t A(2; - 3), B(3; - 2), c di n tch bng 32 v trng tm thuc ng thngA: 3x y 8 = 0. Tm ta nh C. Ta c: AB =2 , M = ( 5 5;2 2 ), pt AB: x y 5 = 0 SABC A= 12d(C, AB).AB = 32 d(C, AB)= 32 GiG(t;3t-8) l trng tm tam gi c ABC thd(G, AB)= 12 d(G, AB)= (3 8) 52t t =12 t = 1 hoc t = 2 G(1; - 5) hoc G(2; - 2) M3 CM GM = C = (-2; 10) hoc C = (1; -4) Cu VI I .a (1 i m) T cc ch s 0,1,2,3,6,7,8,9 c th lp c bao nhiu s t nhin c 6 ch s imt kh c nhau ( ch s u ti n phikh c 0) trong phic ch s 7.Gis c 6 ch s labcdefNu a = 7 thc 7 c ch chn b, 6 c ch chn c, 5 c ch chn d, 4 c ch chn e, 3 c ch chn f . y c 7.6.5.4.3 = 2520s Nu b = 7 thc 6 c ch chn a, 6 c ch chn c, 5 c ch chn d, 4 c ch chn e, 3 c ch chn f . y c 6.6.5.4.3 = 2160s Tng t v ic, d, e, fVy tt c c 2520+5.2160 = 13320 s Cu VI I I .a (1 i m) Tm a bt phng trnh sau c nghim:21 13 3log 1 log ( ) x ax a + > + i u ki n: ax + a > 0 Bpt tng ng 21 ( 1) x ax + < +Nu a>0 thx +1 >0.Ta c 211xax+0 t( )2log3 1x+=u, ( )2log3 1xv =ta c pt u +uv2 = 1 + u2 v2(uv2-1)(u 1) = 0 211uuv=

=. . . x =1 www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 532). Cho cc s thc a, b, c tha mn :0 1, 0 1, 0 1 a b c < s < s < s . Chng minh rng:( )1 1 1 11 3 a b cabc a b c| |+ + + >+ + + |\ . V0 1, 0 1 a b < s < snn( )( ) 1 1 0 1 0 a b ab a b > + >1 a b ab > + ( )1 1 11 1ab a b > + Chng minh tng t :( ) ( )1 1 1 1 1 11 2 , 1 3bc b c ca c a> + > + Cng cc BT (1), (2), (3) v theo v :( )1 1 1 1 1 12 3 4ab bc ca a b c| |+ + > + + |\ . S dng BT (4) v BT Cauchy ta c :( )1 1 1 1 1 1 11 2 3 a b c a b c a b cabc ab bc ca a b c| | | |+ + + = +++ + + > + ++ + + ||\ . \ . ( )1 1 1 1 1 12 3 a b ca b c a b c| |> + + + + + + + |\ . Cng theo BT Cauchy ta c :( )1 1 19 a b ca b c| |+ + + + > |\ . Do ( )1 1 1 1 1 1 11 6 3 3 a b cabc a b c a b c| |+ + + > + + + =+ + + |\ . (pcm) ng thc xy ra khi v ch khi a = b = c = 1. 3). Gii h phng trnh: 3 1 2 322 2 3.23 1 1x y y xx xy x+ + + =+ + = + PT( )( )21 x+1 0 123 1 0 0 1 3 3 1 1 xxx x y x y x x xy x> > > + = = v = + + = + Vi x = 0 thay vo (1) : 228 82 2 3.2 8 2 12.2 2 log11 11y y y y yy+ = + = = =Vi 11 3xy x> = thay y = 1 3x vo (1) ta c :( )3 1 3 12 2 3.2 3x x + + =t 3 12xt+= , v1 x > nn 14t > PT (3) : 23 2 216 6 1 03 2 2tt t ttt

= + = + = = + i chiu iu kin 14t > ta chn3 2 2 t = + .Khi ( )3 1212 3 2 2 log 3 2 2 13xx+ (= + = + ( ) 21 3 2 log 3 2 2 y x = = +www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 54Vy HPT cho c 2 nghim 208log11xy= = v ( )( )221log 3 2 2 132 log 3 2 2xy (= + = + 11 PHN CHUNG CHO TT C CC TH SINH Cu I (2 im) Cho hm s ( ) ( )3 2 2 2y x 3mx 3 m 1 x m 1 = + ( m l tham s)(1). 1.Kho st s bin thin v v th ca hm s (1) khim 0. =2.Tmccgitrcamthhms(1)cttrchonhti3imphnbitc honh dng . Cu II (2 im) 1.Gii phng trnh: 2sin 2x 4sin x 1 0.6t | | + += |\ . 2.Gii h phng trnh: ( )( )( )( )( )2 22 2x y x y 13x, y .x y x y 25 + =e+ =

Cu III (1 im) Cho hnh chpS.ABCD c yABCD l hnh ch nht viAB a, AD 2a, = =cnhSA vung gcviy,cnhSBtovimtphngymtgc o60 . TrncnhSAlyimMsao choa 3AM3= .Mtphng( ) BCM ctcnhSDtiimN.Tnhthtchkhichp S.BCNM.Cu IV (2 im) 1.Tnh tch phn: 62dxI2x 1 4x 1=+ + +} 2.Tm gi tr ln nht v gi tr nh nht ca hm s :y = 2sin8x + cos42x PHN T CHN: Th sinh chn cu V.a hoc cu V.bCu V.a.( 3 im ) Theo chng trnh Chun 1/.Cho ng trn (C) : ( ) ( )2 2x 1 y 3 4 + =v im M(2;4) . a/.Vit phng trnh ng thng i qua M v ct ng trn (C) ti hai im A, B sao cho M l trung im ca AB b/.Vit phng trnh cc tip tuyn ca ng trn (C) c h s gc k = -1 . 2/.Cho hai ng thng song song d1 v d2. Trn ng thng d1 c 10 im phn bit, trn ng thng d2 c n im phn bit ( n 2 > ). Bit rng c 2800 tam gic c nh l cc im cho. Tm n. Cu V.b.( 3 im ) Theo chng trnh Nng cao 2/.p dng khai trin nh thc Niutn ca ( )1002x x + , chng minh rng: 99 100 198 1990 1 99 100100 100 100 1001 1 1 1100C 101C 199C 200C 0.2 2 2 2| | | | | | | | + + = ||||\ . \ . \ . \ . 1/. Cho hai ng trn : (C1) : x2 + y2 4x +2y 4 = 0 v (C2) : x2 + y2 -10x -6y +30 = 0 c tm ln lt l I, J www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 55a/.Chng minh (C1) tip xc ngoi vi (C2) v tm ta tip im H . b/.Gi (d) l mt tip tuyn chung khng i qua H ca (C1) v (C2) . Tm ta giao im K ca (d) v ng thng IJ . Vit phng trnh ng trn (C) i qua K v tip xc vi hai ng trn (C1) v (C2) ti H . P N 11 PHN CHUNG CHO TT C CC TH SINH Cu I (2 im) Cho hm s ( ) ( )3 2 2 2y x 3mx 3 m 1 x m 1 = + ( m l tham s)(1). 1).Kho st s bin thin v v th ca hm s (1) khim 0. =V im = 0 , ta c : y = x3 3x + 1 - TX :R - S bi n thi n: + ) Gi ih n : x xLim y ; Lim y += = ++) Bng bi n thi n: Ta c : y= 3x2 3 y= 0 x = -1 hoc x = 1 Hm s ng bin trn mi khong( ) ; 1 v( ) 1; + , nghch bi n tr n khong ( -1; 1) Hm s t cc i ti im x = -1, gi tr cc i ca hm s l y(-1) =3 Hm s t cc tiu ti im x = 1, gi tr cc tiu ca hm s l y(1) =-1 - th+ i m un : Ta c : y = 6x , y"= 0 t ii m x = 0 v y" i du t dng sang m khi x qua im x = 0 . Vy U(0 ; 1) l i m un ca th. + Gi ao i m v itrc tung :(0 ;1) + THS iqua c c i m : A(2; 3) , B(1/2; -3/8) C(-2; -1) 2.Tm cc gi tr ca m th hm s (1) ct trc honh ti 3 im phn bit c honh dng . THS (1) ct trc honh ti 3 im phn bit c honh dng, ta phi c :y y x+ + + -1 + 00- 1 3 -1 642-2-4-5 5 10 y x www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 56( ) ( )( )1 2y '12x x0x 0x 0y y 0y 0 0>> > + > < < + < < Cu II (2 im) 1/.Gii phng trnh: 2sin 2x 4sin x 1 0.6t | | + += |\ . Ta c :2sin 2x 4sin x 1 0.6t | | + += |\ . 3 si n2x cos2x + 4si nx + 1 = 03 si n2x + 2si n2x + 4 si nx = 0 si nx ( 3cosx + si nx + 2 ) = 0 si nx = 0 (1)hoc3cosx + si nx + 2 = 0 (2) + (1)x = t k+ (2) 3 1cosx si n x 12 2 + = si n x 13t | | + = |\ .5x 26t= + t k2.Gii h phng trnh: ( )( )( )( )( )2 22 2x y x y 13x, y .x y x y 25 + =e+ =

( )( )( )( )( )( )2 22 2x y x y 13 1x y x y 25 2 + =+ = ( )( )3 2 2 33 2 2 3x xy x y y 13 1'y xy x y x 25 2 '+ = + = Ly (2 ) - (1) ta c :x2 yxy2 = 6 ( ) x y xy 6 =(3) Kt h p v i(1) ta c :( )( )( )( )2 2x y x y 13Ix y xy 6 + = = . t y = - z ta c : www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 57( )( )( )( )( ) ( )( )22 2x z x z 13 x z x z 2xz 13Ix z xz 6x z xz 6(+ + = + + = ( + = + = t S = x +z v P = xz ta c :( )23S S 2P 13 S 1 S 2SP 13P 6 SP 6SP 6 = = = = = =

Ta c :x z 1x.z 6+ = = . H ny c nghim x 3z 2= = hoc x 2z 3= = Vy h cho c 2 nghim l : ( 3 ; 2) v ( -2 ; -3 ) Cu III (1 im) Cho hnh chpS.ABCD c yABCD l hnh ch nht viAB a, AD 2a, = =cnhSA vung gc vi y, cnhSB to vi mt phng y mt gc o60 .Trn cnhSA ly imM sao choa 3AM3= .Mt phng ( ) BCMct cnhSD ti imN. Tnh th tch khi chp S.BCNM.Ta c ( SAB)( BCNM) v( ) ( ) SAB BCNM BM = . T S h SH vung gc vi ng thng BM thSH(BCNM)hay SH l ng caoca hnh chp SBCNM. Mt kh c :SA = AB.tan600 = a 3. Suy ra : MA = 13SA Cu IV (2 im) 1/.Tnh tch phn: 62dxI2x 1 4x 1=+ + +} tt 4x 1 = +, ta c dt = 2dx4x 1 + hay t2dt = dx v 2t 1x4=Khi x = 2 th t = 3 v khi x= 6 th t = 5 Khi : 523tdtIt 12 1 t2=| | + + |\ .}=( )523tdtt 1 +}( )5231 1dtt 1t 1| | |= | ++\ .} = 531l n t 1t 1| |+ + |+\ .= 3 1l n2 122/.Tm gi tr ln nht v gi tr nh nht ca hm s :y = 2sin8x + cos42x t t = cos2x( ) 1 t 1 s sth si n2x = 1 t2 NDBCASMH www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 58( ) ( ) ( )3 33 31 1f ' t 4t t 1 8t t 12 2 (= + = + ( ) ( ) ( )2212t t 1 4t 2t t 1 t 12 (= + + + = ( )( )213t 1 7t 4t 12 +Bng bi n thi n Qua bng bi n thi n ta c :mi ny = 127 v maxy = 3 PHN T CHN: Th sinh chn cu V.a hoc cu V.bCu V.a.( 3 im ) Theo chng trnh Chun 1).Cho ng trn (C) : ( ) ( )2 2x 1 y 3 4 + =v im M(2;4) . a).Vit phng trnh ng thng i qua M v ct ng trn ( C) ti hai im A, B sao cho M l trung im ca ABng trn : ( x 1)2 + ( y 3 )2 = 4c tm I( 1 ; 3) v bn knhR = 2 .Ta c :(d) :( ) ( )( )( )Qua M 2; 4qua M qua Md : d :MA MN AB MIvtpt MI 1;1 = (d) : x 2 + y 4 = 0(d) : x + y 6 = 0 b).Vit phng trnh cc tip tuyn ca ng trn I c h s gc k = -1 . ng thng (d) vi h s gc k = -1 c d ng : y = -x + mhay x + y m =0 (1)ng thng (d) l tip tuyn ca ng trn I kc(I ,(d)) = R 121 3 m m 4 2 221 1 m 4 2 2

+ = + = + = + Vy c 2 tip tuyn tho mn bi l :x + y 42 2 = 0 2.Chohaingthngsongsongd1vd2.Trnngthngd1c10imphnbit,trn ng thng d2 c n im phn bit ( n 2 > ). Bit rng c 2800 tam gic c nh l cc im cho. Tm n. Theo ra ta c :3 3 3n 10 10 nC C C 2800+ = (n 2 > ) ( )( ) ( )n 1010! n!28003! n 7 ! 3! 7! 3! n 3 !+ =+ ( )( )( ) ( )( ) n 10 n 9 n 8 10.9.8 n n 1 n 2 2800.6 + + + =n2 + 8n 560 = 0n 20 2n 28= <

=Vy n = 28 Cu V.b.( 3 im ) Theo chng trnh Nng cao t f(t) f(t) -11/31 + 0- 3 127 1 www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 591.p dng khai trin nh thc Niutn ca ( )1002x x + , chng minh rng: 99 100 198 1990 1 99 100100 100 100 1001 1 1 1100C 101C 199C 200C 0.2 2 2 2| | | | | | | | + + = ||||\ . \ . \ . \ . Ta c : [ (x2 + x )100] = 100(x2 + x )99( 2x +1)(1)v ( )1002 0 100 1 101 2 102 99 199 100 200100 100 100 100 100x x C x C x C x C x C x + = + + + + + ( )1002 0 99 1 100 99 198 100 199100 100 100 100x x ' 100C x 101C x 199C x 200C x ( + = + + + + ( (2)T (1) v (2) ta thay 1x2= , ta c99 100 198 1990 1 99 100100 100 100 1001 1 1 1100C 101C 199C 200C 0.2 2 2 2| | | | | | | | + + = ||||\ . \ . \ . \ . 2.. Cho hai ng trn : (C1) : x2 + y2 4x +2y 4 = 0 v (C2) : x2 + y2 -10x -6y +30 = 0 c tm ln lt l I, J a).Chng minh (C1) tip xc ngoi vi (C2) v tm ta tip im H . (C1) c tm I ( 2 ; -1) v bn knh R1= 3 . (C2) c tm J(5;3) v bn knh R=2. Ta c :I J2 = ( 5 2)2 + ( 3 + 1)2 = 25 I J = 5 = R1 + R2

Suy ra (C1) v (C2) tip xc ngoi vi nhau . Ta tip im H c xc nh bi: ( ) ( )( ) ( )HI H J HI H J HH19x2 x x 3 x x52HI 3HJ7 2 y y 3 y yy5= = = = = b).Gi (d) l mt tip tuyn chung khng i qua H ca (C1) v (C2) . Tm ta giao im K ca (d) v ng thng IJ . Vit phng trnh ng trn (C) i qua K v tip xc vi hai ng trn (C1) v (C2) ti H . C :2KI 3KJ = ( ) ( )( ) ( )I K J KKK I K J K2 x x 3 x x x 11y 11 2 y y 3 y y = = = = ng trn (C) qua K , tip xc vi (C1) , (C2) ti H nn tm E ca (C) l trung im ca KH : 37 31E ;5 5| | |\ .. Bn knh (C) l EH = 6 Phng trnh ca (C) l : 237 31x y 365 5| | | | + = ||\ . \ . 12 A. PHN CHUNG CHO CC TH SINH (7im): Cu I: Cho hm s 3 23 3 3 2 y x mx x m = + +(Cm) a) Kho st s bin thin v v th hm s khi m = 13 . b) Tm m (Cm) ct trc honh ti 3 im phnbit c honh l 1 2 3, , x x xtha mn 2 2 21 2 315 x x x + + >www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 60Cu II:a) Gii bt phng trnh: 4log (log (2 4)) 1xx s b) Gii phng trnh: ( )2cos 2 cos 2tan 1 2 x x x + =Cu III: Tnh tch phn : 220I cos cos 2 x xdx=} CuIV:CholngtrngABC.A1B1C1cAB=a,AC=2a,AA12a 5 = v o120 BAC =.. Gi M l trungim ca cnh CC1. Chng minh MB MA1v tnh khong cch d t im A ti mt phng (A1BM). Cu V:Tm m phng trnh sau c mt nghim thc: 22 2( 4) 5 10 3 0 x m x m x + + + + =B. PHN RING (3im): Th sinh ch c lm 1 trong 2 phn Theo chng trnh chun: Cu VI.a:1)Trong mp to (Oxy) cho 2 ng thng: (d1): 7 17 0 x y + = , (d2): 5 0 x y + = . Vit phng trnh ng thng (d) qua im M(0;1) to vi (d1),(d2) mt tam gic cn ti giao im ca (d1),(d2). 2)TrongkhnggianOxyzchohnhhpchnhtABCDABCDcAO,B(3;0;0), D(0;2;0), A(0;0;1). Vit phng trnh mt cu tm C tip xc vi AB. CuVII.a:Mtkschc15quynsch(4quyntonkhcnhau,5quynlkhcnhau,6 quyn vn khc nhau). Ngi ta ly ngu nhin 4 quyn sch t k. Tnh xc sut s sch ly ra khng 3 mn. Theo chng trnh nng cao: Cu VI.b: Trong khng gian Oxyz cho im M(0;1;1)v 2 ng thng: (d1):1 23 2 1x y z += =; (d2) l giao tuyn ca 2 mp c PT:1 0 x + =v2 0 x y z + + =1) Chng t 2 ng thng d1, d2 cho nhau v tnh khong cch gia chng. 2) Vit PT ng thng (d) qua M vung gc (d1) v ct (d2). Cu VII.b: Tm h s ca 8xkhai trin Newtn ca biu thc ( )82 31 P x x = + P N 12 A. PHN CHUNG CHO CC TH SINH (7im): Cu I: Cho hm s 3 23 3 3 2 y x mx x m = + +(Cm) a) Kho st s bin thin v v th hm s khi m = 13 . b) Tm m (Cm) ct trc honh ti 3 im phn bit c honh l 1 2 3, , x x xtha mn 2 2 21 2 315 x x x + + >Phng trnh honh giao im: 3 23 3 3 2 0 x mx x m + + =2 2( 1)[ (3 1) 3 2]=0 1 (3 1) 3 2 0 (2) x x m x m x x m x m = v =(Cm) ct trc honh ti 3 im phn bit c honh l 1 2 3, , x x xvi 31 x =th1 2, x xl nghim khc 1 ca PT (2) Theo l viet ta c:1 21 23 13 2x x mx x m+ = = www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 61 tho mn k th: 2222 2 2 21 2 309 6 9 01 (3 1).1 3 2 0 015 9 9 0m mm m mx x x mA > + + > = = + + > > ( ; 1] [1; ) m e +Cu II:a) Gii bt phng trnh: 4log (log (2 4)) 1xx s4log (log (2 4)) 1xx s . k: 4 20 1log (2 4) 0 log 52 4 0xxxx< = > > > Do1 x > PT4log (2 4) 2 4 4 4 2 4 0x x x x xx s s + >ng vi mi x. Do vy BPT c nghim: 2log 5 x >a)Gii phng trnh: ( )2cos 2 cos 2tan 1 2 x x x + = k:cos 0 / 2 x x k = = +PT 221(2cos 1) cos [2( 1) 1] 2cosx xx + =3 22cos 3cos 3cos 2 0 x x x + =2(cos 1)(2cos 5cos 2) 0 x x x + + =cos 12cos 1/ 22cos 2( ) 3xx kxx kx VN = = +

=

= + = Cu III: Tnh tch phn : 220I cos cos 2 x xdx=} 2 2 220 0 01 1I cos cos 2 (1 cos 2 ) cos 2 (1 2cos 2 cos 4 )2 4x xdx x xdx x x dx = = + = + +} } } /201 1( sin 2 sin 4 ) |4 4 8x x x= + + =Cu IV: Cho lng tr ng ABC.A1B1C1 c AB = a, AC = 2a, AA1 2a 5 = v o120 BAC =..GiMltrungimcacnhCC1. ChngminhMBMA1vtnhkhongcchdtimAtimt phng (A1BM). Theo l cosin ta c: BC =7 aTheo Pitago ta c: MB =2 3a ; MA1=3aVy 2 2 2 21 121 MB MA BA a + = =1MA MB Ta li c: 1 1 111 1( , ( )). .3 3ABA M ABA MBAV dM ABA S d S = =1 1( , ( )) ( , ( )) 3 dM ABA dC ABA a = =1211. 52ABAS AB AA a = =1211. 3 32MBAS MB MA a = =53ad =A1 M C1B1 B A C www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 62Cu V:Tm m phng trnh sau c mt nghim thc: 22 2( 4) 5 10 3 0 x m x m x + + + + =22 2( 4) 5 10 3 0 x m x m x + + + + =22 2( 4) 5 10 3 x m x m x + + + = 2 23 02 2( 4) 5 10 ( 3)xx m x m x> + + + = 232 12 5xx xmx> += Xt hm s, lp BBT vi 22 1( )2 5x xf xx +=222( 5 )'( )(2 5)x xf xx = Khi ta c: Bng bin thin:x- 0 5/2 35+ y+0--0+ y 8 24/5 + Phng trnh c 1 nghim24(8; )5m e+ ` ) B. PHN RING (3im): Th sinh ch c lm 1 trong 2 phn Theo chng trnh chun: Cu VI.a:1)Trong mp to (Oxy) cho 2 ng thng: (d1): 7 17 0 x y + = , (d2): 5 0 x y + = . Vit phng trnh ng thng (d) qua im M(0;1) to vi (d1),(d2) mt tam gic cn ti giao im ca (d1),(d2). Phng trnh ng phn gic gc to bi d1, d2 l: 12 2 2 223 13 0( ) 7 17 53 4 0 ( )1 ( 7) 1 1x y x y x yx y+ = A + + = = A+ + PT ng cn tm i qua M(0;1) v song song vi 1 2, A AKL:3 3 0 x y + =v3 1 0 x y + =2)TrongkhnggianOxyzchohnhhpchnht ABCDABCDcAO,B(3;0;0),D(0;2;0),A(0;0;1).Vit phng trnh mt cu tm C tip xc vi AB. K CHAB, CKDC Ta chng minh c CK (ADCB) nn tam gic CKH vung ti K. 2 2 24910CH CK HK = + =Vy PT mt cu l: 2 2 249( 3) ( 2)10x y z + + =CuVII.a:Mtkschc15quynsch(4quyntonkhcnhau,5quynlkhcnhau,6 quyn vn khc nhau). Ngi ta ly ngu nhin 4 quyn sch t k. Tnh xc sut s sch ly ra khng 3 mn.S phn t ca khng gian mu l: 4151365 C O = =C C D DA B B H K A www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 63Gi A l bin c ly ra 4 quyn sch c 3 mn ( c cc trng hp: (2 ton, 1 l, 1 vn); (1 ton, 2 l, 1 vn);(1 ton, 1l, 2 vn)) c s phn t l: 2 1 1 1 2 1 1 1 24 5 6 4 5 6 4 5 6. . . 720 A C C C CCC CC C = + + =Xc sut xy ra A l: 720 48( ) 0.5271365 91PA = = ~Vy xc sut cn tm l:( )43P 1 P A91= = Theo chng trnh nng cao: Cu VI.b: Trong khng gian Oxyz cho im M(0;1;1)v 2 ng thng: (d1):1 23 2 1x y z += =; (d2) l giao tuyn ca 2 mp c PT:1 0 x + =v2 0 x y z + + =1) Chng t 2 ng thng d1, d2 cho nhau v tnh khong cch gia chng. Ta c: 1Ai qua M1 = (1;-2;0), c vect ch phng 1(3; 2;1) u = Ta tm c 2Ai qua M2 = (-1;-1;0), c vect ch phng 2(0;1;1) u = 1 2 1 2 1 2, (1; 3;3); , . 1 0 u u u u M M (( = = = 1A ,2Acho nhau. 1 2 1 21 2 1 21 2, .1, 19 ( , )19,u u M Mu u dd du u ( ( = = = ( 2) Vit PT ng thng (d) qua M vung gc (d1) v ct (d2). Mp(P) i qua M(0;1;1) vung gc vi d1 c PT:3 2 3 0 x y z + + =Giao im A ca d2 v (P) l nghim ca h 3 2 3 0 11 0 5 / 32 0 8 / 3x y z xx yx y z z+ + = = + = = + + = = T cn tm l AM c PT: 1 13 2 5x y z = =Cu VII.b: Tm h s ca 8xkhai trin Newtn ca biu thc ( )82 31 P x x = + Ta c: ( )882 2801 (1 ) (1 )k k kkP x x Cx x== + = . M 0(1 ) ( 1)kk i i ikix C x= = ng vi 8xta c:2 8; 0 8 0 4 k i i k k + = s s s s sTa c:k01234 i86420 LoiLoiLoiTMTM Do vy h s ca 8xl: 3 2 2 4 0 08 3 8 4( 1) ( 1) 238 a CC CC = + = 13 I:PHN CHUNG CHO TT C CC TH SINH (7,0 im) Cu I (2.0 im).Cho hm s 4 25 4, y x x = +c th (C) 1. Kho st v v th (C). www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 642. Tm m phng trnh 4 22| 5 4 | log x x m + =c 6 nghim. Cu II (2.0 im). 1. Gii phng trnh: 1 1sin 2x sin x 2cot 2x2sin x sin 2x+ =2. Tm m phng trnh: ( )2m x 2x 2 1 x(2 x) 0 (2) + + + sc nghim x 0; 1 3 (e + Cu III (1.0 im).Tnh 402x 1I dx1 2x 1+=+ +} Cu IV (2.0 im).1.Cho lng tr ng ABCA1B1C1 c AB = a, AC = 2a, AA12a 5 =v o120 BAC =.. Gi M l trung im ca cnh CC1. a/.Chng minh MBMA1

b/.Tnh khong cch d t im A ti mt phng (A1BM). II. PHN RING (3.0 im) 1)Cu VI.a. (2.0 im). 1. Trong khng gian Oxyz cho hai im A (-1; 3; -2), B (-3; 7; -18) v mt phng (P): 2x - y + z + 1 = 0 a. Vit phng trnh mt phng cha AB v vung gc vi mp (P). b. Tm ta im M e (P) sao cho MA + MB nh nht. 2. (1.0 im). Giiphng trnh: ( )2 23 3log 1 log 2 x x x x x + + = 2)Cu V.b. (1,5im). 1.Gii bt phng trnh: 2x 4 2(log 8 log x ) log 2x 0 + >2.(1.5 im).Cho x, y, z l cc s dng. Chng minh : 3 2 4 3 5 x y z xy yz zx + + > + + P N 13 I:PHN CHUNG CHO TT C CC TH SINH (7,0 im) Cu I (2.0 im).Cho hm s 4 25 4, y x x = +c th (C) 1. Kho st v v th (C). 2. Tm m phng trnh 4 22| 5 4 | log x x m + =c 6 nghim.

944129log 12 144 124m m = = =Cu II (2.0 im). 1. Gii phng trnh: 1 1sin 2x sin x 2cot 2x2sin x sin 2x+ = (1) www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C P NN THII HC 65(1) cos22x cosxcos2x = 2cos2x v sin2x = 0 = + + =2cos2x 0v 2cos x cosx 1 0(VN) cos2x = 0 t t t= + t = + 2x k x k2 4 2 2.Tm m phng trnh: ( )2m x 2x 2 1 x(2 x) 0 (2) + + + sc nghim x0; 1 3 (e + t 2t x 2x 2 = + t2 2 = x2 2x Bpt (2) s s s e ++2t 2m (1 t 2), dox [ 0;1 3]t 1 Kho st 2t 2g(t)t 1=+vi1 s t s 2 g'(t) 22t 2t 20(t 1)+ += >+. Vy g tng trn [1,2] Do , ycbtbpt 2t 2mt 1s+ c nghim t e [1,2] | | es = =t 1;22m max g(t) g(2)3 Vy ms23 Cu III (1.0 im).Tnh 402x 1I dx1 2x 1+=+ +} t 2t 2x 1 t 2x 1 2tdt 2dx dx tdt = + = + = = ;i cn t(4) = 3, t(0) = 1Vy 4 3 320 1 12x 1 t 1I dx dt t 1 dt1 t t 1 1 2x 1+ | |= = = + |+ + + + \ .} } };= Cu IV (2.0 im).1.Cho lng tr ng ABCA1B1C1 c AB = a, AC = 2a, AA12a 5 =v o120 BAC =.. Gi M l trung im ca cnh CC1. a).Chng minh MBMA1

Chn h trc Axyz sao cho: A 0,( ) C 2a, 0, 0 , 1A (0, 0, 2a 5) | | | |\ .a a 3A(0; 0; 0), B ; ; 02 2 v M( 2a, 0, a 5) | | = = | |\ . 15 3BM a ; ; 5 , MA a(2; 0; 5)2 2 Ta c:= + + = 21 1BM.MA a ( 5 0 5) 0 BM MAb.Tnh khong cch d t im A ti mt phng (A1BM). Ta c th tch khi t din AA1BM l : www.VIETMATHS.comwww.VIETMATHS.com www.VIETMATHS.comB TON C