Bo de Toan Thi Thu Dh Co Giai Chi Tiethay

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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: