25 Đề Thi Thử Đại Học Môn Toán 2015 Có Lời Giải Và Đáp Án Chi Tiết

B GIO DC V O TO

THI TH

THI TUYN SINH I HC, CAO NG Mn thi: TON; Khi A v khi A1

Thi gian lm bi: 180 pht, khng k thi gian pht

PHN CHUNG CHO TT C TH SINH (7,0 im).

Cu 1 (2,0 im). Cho hm s 2 3

.1

xy

x

a) Kho st s bin thin v v th (C) ca hm s cho.

b) Tm m ng thng 2y x m ct th (C) ti hai im phn bit c tung dng.

Cu 2 (1,0 im). Gii phng trnh 3

(tan 2 cot 1)sin 4 sin 2sin cos .3 2 2

x xx x x x

Cu 3 (1,0 im). Gii bt phng trnh 2

2

62 1 1.

( 2 1 1)

xx x

x

Cu 4 (1,0 im). Tnh din tch hnh phng gii hn bi cc ng (1 ) ;xy x e

3 1y x v trc tung.

Cu 5 (1,0 im). Cho hnh chp u S.ABC c gc gia mt bn v mt y bng 060 , khong cch gia hai ng

thng SA v BC bng 3

.2 7

a Tnh theo a th tch khi chp S.ABC v din tch mt cu i qua bn im S, O, B, C vi

O l tm y.

Cu 6 (1,0 im). Cho ba s thc dng , ,a b c tha mn 1.abc Tm gi tr ln nht ca biu thc

2 2 2

1 1 1.

5 5 5A

a ab a b bc b c ca c

PHN RING (3,0 im): Th sinh ch c lm mt trong hai phn (phn A hoc phn B)

A. Theo chng trnh Chun

Cu 7.a (1,0 im). Trong mt phng vi h ta Oxy, cho ng trn (C): 2 2( 1) ( 2) 5x y v im A (2;0).

Tm ta hai im B, C thuc (C) sao cho tam gic ABC vung ti B v c din tch bng 24.

Cu 8.a (1,0 im). Trong khng gian vi h ta Oxyz, cho tam gic u ABC c A (4;2; 6) v phng trnh

ng thng BC l 3 3 1

.2 1 1

x y z Vit phng trnh ng thng d i qua trc tm tam gic ABC v vung

gc vi mt phng (ABC).

Cu 9.a (1,0 im). Tm s hng khng cha x trong khai trin nh thc Niu-tn ca 23

2

n

xx

( 0x ) bit rng

1 2 32 3 ... ... 256 .k nn n n n nC C C kC nC n

B. Theo chng trnh Nng cao

Cu 7.b (1,0 im). Trong mt phng vi h ta Oxy, vit phng trnh chnh tc ca elip (E), bit rng khi M thay

i trn (E) th di nh nht ca OM bng 4 v di ln nht ca 1MF bng 8, 1F l tiu im c honh m.

Cu 8.b (1,0 im). Trong khng gian vi h ta Oxyz, cho ng thng 1 1

:2 1 1

x y zd v mt phng

( ) : 1 0.P x y z Vit phng trnh ng thng thuc mt phng (P) sao cho vung gc vi d v

khong cch gia v d bng 3.

Cu 9.b (1,0 im). Tm s phc z bit 2 2z z l s thc v

1z

z c mt acgumen bng .

3

------------ Ht -------------

B GIO DC V O TO

THI TH



PHN CHUNG CHO TT C TH SINH (7,0 im)

Cu 1 (2,0 im). Cho hm s 1

x my

x ( 1)m (1)

a) Kho st s bin thin v v th hm s (1) khi 1.m

b) Gi 1k l h s gc ca tip tuyn ti giao im ca th hm s (1) vi trc honh. Gi 2k l h s gc ca

tip tuyn vi th hm s (1) ti im c honh 1.x Tm tt c cc gi tr ca tham s m sao cho

1 2k k t gi tr nh nht.

Cu 2 (1,0 im). Gii phng trnh

2

2

sin3 cos tan 2tan 12 .

cos2 1 tan

x x x x

x x

Cu 3 (1,0 im). Gii bt phng trnh

236 2 5

3

xx x

x ( x ).

Cu 4 (1,0 im). Tnh din tch hnh phng gii hn bi cc ng 24 3y x v 2 .y x

Cu 5 (1,0 im). Cho hnh lng tr ng ABC.ABC c y ABC l tam gic vung ti A, AB 2 ;a AA .a Gc

gia mt phng (ABC) v (ABC) bng 030 . Tnh th tch ca khi lng tr ABC.ABC v khong cch gia hai

ng thng BC v AC theo .a

Cu 6 (1,0 im). Cho cc s thc dng zyx ,, tha mn iu kin .1xyz Tm gi tr nh nht ca biu thc

.)1(3

4

)1(

1

)1(

1322 zyx

P

PHN RING (3,0 im): Th sinh ch c lm mt trong hai phn (phn A hoc phn B) A. Theo chng trnh Chun

Cu 7.a (1,0 im). Trong mt phng vi h ta Oxy, cho hnh bnh hnh ABCD c nh D ( 6; 6), ng trung

trc ca on thng CD c phng trnh l 2 3 17 0x y v ng phn gic ca BAC c phng trnh l 5 3 0.x y Xc nh ta cc nh cn li ca hnh bnh hnh.

Cu 8.a (1,0 im). Trong khng gian vi h ta Oxyz, cho im A ( 1;2;0) v hai ng thng

1

1 3: ,

1 3 4

x y z 2

2: .

2 1 2

x y z Vit phng trnh mt phng (P) song song vi hai ng thng 1,

2 v cch A mt khong bng 3.

Cu 9.a (1,0 im). Cho s phc z tha mn 4 (1 3 ) 25 21 .z i z i Tnh mun s phc 21 .w z z


Cu 7.b (1,0 im). Trong mt phng vi h ta ,Oxy cho ng trn 2 2( ) : 2 4 20 0C x y x y v hai

ng thng 1 : 2 5 0,d x y 2 : 2 0.d x y Vit phng trnh ng thng tip xc vi ( )C ti ,A ct 1,d

2d ln lt ti B v C sao cho B l trung im ca on thng .AC

Cu 8.b (1,0 im). Trong khng gian vi h ta ,Oxyz cho ng thng 1 2

:2 1 1

x y zd v mt phng

( ) : 2 3 0.P x y z Vit phng trnh ng thng nm trong ( ),P sao cho vung gc vi d v khong

cch gia hai ng thng v d bng 2.

Cu 9.b (1,0 im). Tm s phc z tha 3 3 3z i v c acgumen nh nht vi (0; ). .

------------ Ht ------------

B GIO DC V O TO

THI TH




Cu 1 (2,0 im). Cho hm s 3 22

( 2) 23

y x mx m x (1), vi m l tham s thc.


b) Tm cc gi tr ca m hm s (1) nghch bin trn na khong [0; ).


3 cos 12 3 tan cot .

sin cos

xx x

x x

Cu 3 (1,0 im). Gii h phng trnh 12 3 4 16

4 5 5 6

x y xy

x y ( ,x y ).

Cu 4 (1,0 im). Tnh tch phn .)1(

)2ln(11

0

2x

xI

Cu 5 (1,0 im). Cho hnh lng tr tam gic u ABC.ABC c cnh y bng ,a cnh bn bng 2 .a Gi M, N, E

ln lt l trung im ca cc cnh AC, AA v AB. Mt phng (MNE) ct cnh BC ti F. Tnh (theo a ) th tch ca

khi chp B.MNEF.

Cu 6 (1,0 im). Cho , ,a b c l cc s thc dng. Tm gi tr ln nht ca biu thc

3 33

1 1.

( 1)( 1)( 1)4( ) 1P

a b ca b c



Cu 7.a (1,0 im). Trong mt phng vi h ta Oxy, gi 1 2,d d l cc ng thng qua A ( 2; 3) v hp vi

ng thng :3 17 0x y mt gc 045 . Vit phng trnh ng trn c tm nm trn ng thng

: 15 0d x y ng thi tip xc vi 1d v 2.d

Cu 8.a (1,0 im). Trong khng gian vi h ta Oxyz, cho hnh ch nht ABCD vi A (1;2; 1), im D nm trn

ng thng 2 1

: ,1 2 1

x y z im C nm trn mt phng (P): 3 2 7 0x y z ng thi AC vung gc

vi mt phng (P). Tm ta cc im B, C, D ca hnh ch nht.

Cu 9.a (1,0 im). Vi n l s nguyn dng, gi 3 3na l h s ca 3 3nx trong khai trin thnh a thc ca biu

thc 3 2( ) ( 3 3 9) .nP x x x x Hy tm s n bit rng 3 3 495.na

B. Theo chng trnh nng cao

Cu 7.b (1,0 im). Trong mt phng vi h ta Oxy, cho ng trn (C): 2 2( 1) ( 2) 16x y v ng thng

: 4 3 15 0.d x y Gi H l hnh chiu vung gc ca tm I ng trn (C) ln ng thng .d T im M bt k

trn d k cc tip tuyn MP, MQ n (C) (P v Q l cc tip im). Dy PQ ct IH ti K. Chng minh im K c nh

khi M thay i. Tm ta im K.


:1 2 1

x y zd v mt phng

( ) : 2 2 1 0.x y z Vit phng trnh mt phng (P) cha d v to vi ( ) mt gc nh nht.

Cu 9.b (1,0 im). Gi z l nghim phng trnh 1 3 1 2 3

1 2 3(1 )

i i

z i i i. Vit dng lng gic ca s phc

4.z

------------Ht------------

B GIO DC V O TO

THI TH




Cu 1 (2,0 im). Cho hm s 3 2 22 2y x mx m x (1), vi m l tham s thc.

a) Kho st s bin thin v v th hm s (1) khi 3.m b) Tm m tip tuyn ca th hm s (1) ti im c honh 1x ct cc trc Ox, Oy ln lt ti A, B sao

cho din tch tam gic OAB bng 8

.3

Cu 2 (1,0 im). Gii phng trnh 23cot sin 2 3 cos sin 3(1 3 cot ).x x x x x

Cu 3 (1,0 im). Gii h phng trnh

2 2

2 2

2 2 2

1( ) 1 5

( 1) 2

x yx y

xy x y

( ,x y ).

Cu 4 (1,0 im). Tnh tch phn

1

3

0

3 3 5ln .

( 1) 1

x xI dx

x x

Cu 5 (1,0 im). Cho hnh chp S.ABCD c y ABCD l hnh thang vung ti A v B, 045 ;BCD y ABCD ngoi tip ng trn bn knh bng .a Mt bn SCD l tam gic cn ti S v nm trong mt phng vung gc vi mt

phng (ABCD), cnh bn SB to vi y gc 045 . Tnh (theo a) th tch khi chp S.ABCD.

Cu 6 (1,0 im). Cho cc s thc dng , ,a b c tha mn iu kin 2 2 2( ) 4 4.a b a b c c Tm gi tr nh

nht ca biu thc 2 2 2( ) ( ) 8 7

.( )

a b c b a c c cP

a c b c c a b

PHN RING (3,0 im): Th sinh ch c lm mt trong hai phn (phn A hoc phn B) A Theo chng trnh Chun

Cu 7.a (1,0 im). Trong mt phng vi h ta Oxy, cho tam gic ABC vi B ( 2;1), ng phn gic trong AD

v trung tuyn CM ln lt c phng trnh 1 0x y v 7 5 11 0.x y Vit phng trnh ng trn ngoi

tip tam gic ABC.

Cu 8.a (1,0 im). Trong khng gian vi h ta Oxyz, cho mt phng (P): 6 0x y z v hai im

(1; 1;2),A B (2;1;4). Gi C l im c tung dng nm trn (P) sao cho tam gic ABC vung cn ti B. Vit

phng trnh ng thng l hnh chiu song song ca ng thng AB ln mt phng (P) theo phng BC.

Cu 9.a (1,0 im). Xc nh x trong khai trin nh thc Niu-tn ca 2

4 ,4

n

x

x bit rng tng ca s hng th t

v s hng th su bng 2240 v 1 23 80n nC A ( n ).


Cu 7.b (1,0 im). Trong mt phng vi h ta Oxy, cho ng trn (C): 2 2 16 2 40 0x y x y v ng

thng : 8 0.x y Gi M l im thuc . Qua M k cc tip tuyn MA v MB n (C) (A v B l cc tip im).

Tm ta im M, bit bn knh ng trn ni tip tam gic MAB bng 40

.13

Cu 8.b (1,0 im). Trong khng gian vi h ta Oxyz, cho cc im A (3;1;0), B7

;14;0 ,2

C7

0;0;2

v

ng thng 1 3 4

: .1 1 1

x y zd Gi H l trc tm ca tam gic ABC. Vit phng trnh mt phng (P) cha d

sao cho khong cch t H n (P) l ln nht.

Cu 9.b (1,0 im). Tm s phc ,z bit 2 3z z i v 2 3 3 (3 2 3)

(2 3 )

i

i z c mt acgumen bng .

2

---------- Ht ----------

B GIO DC V O TO

THI TH




Cu 1 (2,0 im). Cho hm s 4 21 1

2 2y x mx (1), vi m l tham s thc.

a) Kho st s bin thin v v th hm s (1) khi 1.m b) Tm m th hm s (1) c ba im cc tr A, B, C sao cho ng trn ngoi tip tam gic ABC c bn

knh bng 3

.2

Cu 2 (1,0 im). Gii phng trnh 3 2

2

3(1 sin 2 )6cot 2cot 8sin .

sin 4

xx x x

x


2 3 2 2 3 3 2 15

( 3 2 ) 2 y 4

x y xy y

x x y ( ,x y ).


2

0

1.

1 2 1 4

xI dx

x x

Cu 5 (1,0 im). Cho hnh chp t gic u S.ABCD c y l hnh vung cnh .a Gi M, N ln lt l trung im

cc cnh SA v BC. Bit gc gia ng thng MN v mt phng (ABCD) bng 060 . Tnh th tch khi chp S.ABCD

v tnh khong cch gia hai ng thng MN v BD theo .a

Cu 6 (1,0 im). Cho zyx ,, l cc s thc dng tha mn .2

3zyx Tm gi tr nh nht ca biu thc

P .141414

222222

xy

xzxz

zx

zyzy

yz

yxyx


Cu 7.a (1,0 im). Trong mt phng vi h ta Oxy, cho ng trn (C): 2 2 6 2 15 0x y x y v im

(3;9).A Qua A k cc tip tuyn AB, AC n ng trn (C) (B v C l cc tip im). Vit phng trnh ng trn

ni tip tam gic ABC.

Cu 8.a (1,0 im). Trong khng gian vi h ta Oxyz, cho im M ( 5; 1;4) v hai ng thng c phng trnh

1

1 1 1: ;

2 2 1

x y zd 2

1 3: .

2 1 3

x y zd Chng minh rng 1d v 2d ct nhau ti im I. Vit phng trnh

ng thng i qua M v ct 1 2,d d ln lt ti A, B khc I sao cho IA IB.

Cu 9.a (1,0 im). Gii bt phng trnh 2 3 3

2 2 2

3log ( 3) 3 log ( 7) log (5 ) .

2x x x


Cu 7.b (1,0 im). Trong mt phng vi h ta Oxy, cho elip (E): 2 2

118 8

x y v hai im A (3;4), B (6;2). Gi

C l im trn (E) sao cho din tch tam gic ABC nh nht. Tnh gi tr ca 2 2sin sin .H CAB CBA Cu 8.b (1,0 im). Trong khng gian vi h ta Oxyz, cho mt phng (P): 5 4 2 0x y z v hai ng

thng 12 2 3

: ,2 1 2

x y z 2

4 2 3: .

1 3 1

x z z Gi A v B l hai im ln lt trn 1 v 2 sao cho

AB vung gc vi (P). Vit phng trnh ng phn gic ACB ca tam gic ABC, bit im C (1;1;2). Cu 9.b (1,0 im). Tm tp hp cc im M trong mt phng phc biu din cc s phc z tha mn iu kin

(1 2 ) 1,z i w trong w v 1

5.2

w i

i

------------ Ht ------------

B GIO DC V O TO

THI TH

THI TUYN SINH I HC, CAO NG

Mn thi: TON; Khi A v khi A1



Cu 1 (2,0 im). Cho hm s mxmxxy )4(3 23 (1), vi m l tham s thc.

a) Kho st s bin thin v v th ca hm s (1) khi .4m

b) Tm m th ca hm s (1) ct trc honh ti 3 im phn bit A ),0;1( B, C sao cho

,011

CB

Akk

k trong CBA kkk ,, ln lt l cc h s gc tip tuyn ca th ti A, B, C.

Cu 2 (1,0 im). Gii phng trnh .3cos)sin23(

)sin25)(sin1(

xx

xx

Cu 3 (1,0 im). Gii phng trnh .02951073823 xx

Cu 4 (1,0 im). Tnh din tch hnh phng gii hn bi cc ng x

xxy

5

54 2

v .4 xy

Cu 5 (1,0 im). Cho hnh chp S.ABC c y ABC l tam gic u, mt bn SAB nm trong mt phng vung gc

vi mt phng y v tam gic SAB vung ti S, SA ,3a SB .a Gi K l trung im ca on AC. Tnh th tch

ca khi chp S.ABC v khong cch gia hai ng thng BC v SK theo .a

Cu 6 (1,0 im). Cho cba ,, l cc s thc dng. Tm gi tr nh nht ca biu thc

.1

16

)2)(2(

9

222 cbacbcaabP



Cu 7.a (1,0 im). Trong mt phng vi h ta Oxy, cho tam gic ABC c trc tm H ( 6;7), tm ng trn

ngoi tip I (1;1) v D (0;4) l hnh chiu vung gc ca A ln ng thng BC. Tm ta nh A.

Cu 8.a (1,0 im). Trong khng gian vi h ta Oxyz, cho ng thng 1

:1 2 2

x y zd v im A (3; 1;2).

Vit phng trnh mt phng (P) vung gc vi d v khong cch t A n (P) bng 2.

Cu 9.a (1,0 im). Tm hai s thc b v c bit iz 1 l nghim ca phng trnh .02 cbzz Khi tnh

mun ca s phc ).12)(12( 21 izizw


Cu 7.b (1,0 im). Trong mt phng vi h ta Oxy, cho elip (E): .134

22 yx Hai im M ),;2( m N );2( n di

ng v tha mn tch khong cch t hai tiu im 21 , FF ca (E) n ng thng MN bng .3 Tnh

1cos .MF N

Cu 8.b (1,0 im). Trong khng gian vi h toa Oxyz, cho ng thng 2 2 1

:1 2 2

x y z v hai im

( 1;1;2),A B (1;0;1). Vit phng trnh ng thng d i qua im B, ct ng thng sao cho khong cch t A

n ng thng d l ln nht, nh nht.

Cu 9.b (1,0 im). Tm s phc z c mun ln nht, nh nht tha mn iu kin 3 3

2.1 2

z i

z i

---------- HT -----------

B GIO DC V O TO

THI TH




Cu 1 (2,0 im). Cho hm s 3 22 ( 2) 1y x mx m x m (1), vi m l tham s thc.

a) Kho st s bin thin v v th hm s (1) khi 1

.3

m

b) Tm m th hm s (1) ct trc honh ti ba im phn bit A (1;0), B, C sao cho BC 6.

Cu 2 (1,0 im). Gii phng trnh cot 2 2cos .4

x x


2

2 2 2

16 2 (1 ) 2 3 0

2 ( ) 1 ( )

x xy y y y

xy x y x y ( ,x y ).


2

2

1

(1 ln ) ln.

( 1)

x x xI dx

x x

Cu 5 (1,0 im). Cho hnh chp S.ABCD c y ABCD l hnh vung cnh .a Cnh bn SA vung gc vi mt phng

(ABCD). Mt phng (SBD) to vi mt phng (ABCD) gc 060 . Gi M l trung im ca AD. Tnh th tch khi chp

S.ABCD v khong cch gia hai ng thng BM v SC theo .a

Cu 6 (1,0 im). Cho , ,a b c l cc s thc dng tha mn 1.a b c Tm gi tr nh nht ca biu thc

2 2 2

2 2

4.

1 1 2 2 2

a b cP

b a a b



Cu 7 (1,0 im). Trong mt phng vi h ta Oxy, cho hnh ch nht ABCD c phng trnh ng thng AB, AC

ln lt l 5 0x y v 3 7 0.x y Trng tm G ca tam gic ACD nm trn ng thng 2 6 0.x y

Tm ta cc nh ca hnh ch nht ABCD.

Cu 8.a (1,0 im). Trong khng gian vi h ta Oxyz, cho ng thng 1 1

:2 1 1

x y zd v im A (1;2;1).

Gi (P) l mt phng i qua A v .d Vit phng trnh mt cu (S) c tm I nm trn mt phng (Oxy) v tip xc vi

mt phng (P) ti A.

Cu 9.a (1,0 im). Gi S l tp hp cc s t nhin c bn ch s phn bit c chn t cc ch s 1,2,3,4,5,6.

Xc nh s phn t ca tp hp S. Chn ngu nhin mt s t tp S, tnh xc sut s c chn c hai ch s cui

l hai ch s chn.


Cu 7.b (1,0 im). Trong mt phng vi h ta Oxy, cho ng trn (C): 2 2 2 4 1 0x y x y v im

(4;1).A Vit phng trnh ng thng d ct (C) ti hai im B, C sao cho tam gic ABC u.


:1 2 3

x y zd v mt phng

( ) : 1 0.P x y z Vit phng trnh chnh tc ca ng thng nm trong (P), vung gc vi d v ct mt

phng (Oxy) ti im M sao cho OM 5.

Cu 9.b (1,0 im). Gii h phng trnh

2 2

2 2

2 2 1 0.

log ( ) log ( 2) 1

x xy y x y

x y y

-------------- Ht ------------

B GIO DC V O TO

THI TH

THI TUYN SINH I HC, CAO NG Mn thi: TON; Khi A v Khi A1

Thi gian lm bi 180 pht, khng k thi gian pht



2

mx my

x c th l ( ),mC vi m l tham s thc.


b) Tip tuyn vi ( )mC ti im c honh 1x ct hai ng tim cn ca ( )mC ln lt ti A v B. Tm

m tam gic OAB c din tch bng 1 (O l gc ta ).

Cu 2 (1,0 im). Gii phng trnh 3 sin

tan 2.2 1 cos

xx

x

Cu 3 (1,0 im). Gii h phng trnh 0853

01)2(1)2(

224

22

yyxyx

xyyx ( yx, ).

Cu 4 (1,0 im). Tnh tch phn 2

cos

0

sin 2 ( 1 3sin ) .xI x x e dx

Cu 5 (1,0 im). Cho hnh chp S.ABCD c y ABCD l hnh thoi cnh bng ,a 0120 ,ABC cnh SA vung gc

vi y (ABCD). Bit gc gia hai mt phng (SBC) v (ABCD) bng 045 . Tnh th tch ca hnh chp S.ABCD v

khong cch gia hai ng thng SC, BD theo .a

Cu 6 (1,0 im). Cho , ,a b c l cc s thc dng v tha mn iu kin 2( )( ) 4 .a b a c a Tm gi tr nh nht

ca biu thc 2 2

2 34 7 .

b c bc bcP

a c a b a a


Cu 7.a (1,0 im). Trong mt phng vi h ta Oxy, cho ng thng : 3 0d y v hai im A (1;2), (3;4).B

Vit phng trnh ng trn (C) i qua hai im A, B v ct ng thng d ti hai im phn bit M, N sao cho

060 .MAN


:2 1 1

x y z v hai im

(4; 4;5),A B (2;0; 1). Tm ta im M thuc ng thng sao cho 2 22 36.MA MB

Cu 9.a (1,0 im). Cho n l s nguyn dng tha mn 2 3 22( ) 3 5 .n nC C n n Tm s hng khng ph thuc x

trong khai trin nh thc Niu-tn ca 34

1,

n

xx

0.x


Cu 7.b (1,0 im). Trong mt phng vi h ta Oxy, cho hnh vung ABCD c nh A (1;1), AB 4. Gi M l

trung im ca cnh BC, K9 3

;5 5

l hnh chiu vung gc ca D ln AM. Tm ta cc nh cn li ca hnh

vung ABCD, bit honh nh B nh hn 2. Cu 8.b (1,0 im). Trong khng gian vi h ta Oxyz, cho tam gic ABC c cc nh A (1;1;1), B (5;1; 2) v

( ; ;1)C a b ( 0, 0).a b Tm ,a b sao cho 12

cos25

BAC v din tch tam gic ABC bng 481.

Cu 9.b (1,0 im). Cho s phc z c phn thc b hn phn o, tng phn thc v phn o bng 3 v mun ca 2014z bng 10075 . Tnh mun ca s phc ( 5).z z

------------ Ht ------------

B GIO DC V O TO

THI TH




Cu 1 (2,0 im). Cho hm s 3 2 2y x mx (1), vi m l tham s thc.


b) Tm m ng thng 2 1y mx m ct th hm s (1) ti ba im phn bit I ( 1;1 ),m A v B

ng thi cc tip tuyn ca th hm s (1) ti A v B c cng h s gc.

Cu 2 (1,0 im). Gii phng trnh 2 2(1 sin )cos (1 cos )sin

1.1 sin 2

x x x x

x


2 2

2

2

1 1

x y x y x xy y

x x y x ( ,x y ).


ln3

0

1.

1

x

x

eI dx

e

Cu 5 (1,0 im). Cho hnh chp S.ABCD c y ABCD l hnh ch nht, AD ,a AB 2,a mt phng (SAB)

vung gc vi mt phng y, cc ng SC, SD cng to vi mt phng (ABCD) nhng gc bng 060 . Gi G l trng

tm ca tam gic SAB. Tnh theo a th tch khi chp S.ABCD v khong cch gia hai ng thng AC v DG.

Cu 6 (1,0 im). Cho cc s thc , , [ 1;1]x y z v tha mn 0.x y z Tm gi tr nh nht ca biu thc

2 2 2 2 2 2

1 1 1.

33 3x xy y y yz z z zx z

P


Cu 7.a (1,0 im). Trong mt phng ta Oxy, cho hnh ch nht ABCD c D (7; 3) v BC 2 AB. Gi M v N

ln lt l trung im ca cc cnh AB v BC. Tm ta nh C, bit phng trnh ng thng MN l

3 16 0.x y

Cu 8.a (1,0 im). Trong khng gian vi h ta Oxyz, cho mt cu (S): 2 2 2 2 2 4 3 0x y z x y z v

mt phng (P): 2 2 0.x y z Vit phng trnh mt phng ( ) i qua im M (1;3;3), vung gc vi (P) v tip

xc vi (S).

Cu 9.a (1,0 im). Cho s phc z tha mn iu kin 10

4 3 .1

zi

i z Tnh

2.w z z


Cu 7.b (1,0 im). Trong mt phng vi h ta Oxy, cho hai ng trn 2 2

1( ) : 2 4 20 0C x y x y v

2 2

2( ) : ( 7) ( 10) 25C x y tip xc vi nhau ti im A (4;6). Tm ta im B thuc 1( ),C im C thuc

2( )C sao cho tam gic ABC vung ti A v khong cch t A n BC l ln nht.

Cu 8.b (1,0 im). Trong khng gian vi h ta Oxyz, cho cc im A ),3;0;0( B )0;1;0( v C ).0;0;2( Gi H l

trc tm ca tam gic ABC. Vit phng trnh mt cu (S) c tm l H v tip xc vi trc Ox.

Cu 9.b (1,0 im). Tm s hng cha 6x trong khai trin nh thc Niu-tn ca 3

12 ,

n

xx

bit rng n l s

nguyn dng tha mn 2 3

0 1 2 32 2 2 1093... .3 4 1 1

nn

n n n n nC C C C Cn n

------------ Ht ------------

B GIO DC V O TO

THI TH




Cu 1 (2,0 im). Cho hm s 3 26 9 2y x mx x m (1), vi m l tham s thc.

a) Kho st s bin thin v v th hm s (1) khi 1.m b) Tm m th hm s (1) c hai im cc tr sao cho ng thng i qua hai im cc tr to vi hai trc ta

mt tam gic cn.

Cu 2 (1,0 im). Gii phng trnh 4 41 (sin cos )

cos .sin

x xx

x


2

2 2

4 4 2 1 4 4 2 1

2

x x y y y

x y x y ( ,x y ).


4

2.

ln 1 3ln

e

e

dxI

x x x

Cu 5 (1,0 im). Cho hnh chp S.ABCD c y ABCD l hnh vung cnh ,a SD 2 5

,3

a hnh chiu vung gc

ca S trn mt phng (ABCD) l trng tm ca tam gic ABD. Tnh theo a th tch ca khi chp S.ABCD v khong cch t A n mt phng (SBC).

Cu 6 (1,0 im). Cho cc s dng ,a b phn bit tha mn 2 2 7.a b Tm gi tr nh nht ca biu thc

4 4 2

2 2 3.

( )P

a b a b


Cu 7.a (1,0 im). Trong mt phng vi h ta Oxy, cho ng trn (C): 2 2 4 2 0x y x y v ng thng

: 2 0.x y Gi I l tm ca (C), M l im thuc . Qua M k cc tip tuyn MA v MB n (C) (A v B l cc

tip im). Vit phng trnh ng trn ngoi tip tam gic MAB, bit din tch t gic MAIB bng 10.

Cu 8.a (1,0 im). Trong khng gian vi h ta Oxyz, cho ng thng 1 2 1

:2 1 3

x y z v im

(2; 5; 6).A Tm ta hnh chiu vung gc ca A trn . Vit phng trnh chnh tc ng thng d i qua A v

ct ti im B sao cho AB 35, im B c honh ln hn 2.

Cu 9.a (1,0 im). Cho s phc z tha mn 20

1 3 .z iz

Tnh mun ca s phc .z


Cu 7.b (1,0 im). Trong mt phng vi h ta Oxy, cho elip (E): 2 2

125 9

x y v im C (5;0). Tm ta cc

im A v B thuc (E) sao cho tam gic ABC cn ti C v c din tch ln nht.

Cu 8.b (1,0 im). Trong khng gian ta Oxyz, cho hai ng thng 2 1

:1 1 1

x y zd v

1 4

: .

1 2

x t

y t

z t

Vit phng trnh mt phng (P) song song vi hai ng thng ,d v ct cc trc ta ln lt ti A, B, C sao

cho t din OABC c th tch bng 6.

Cu 9.b (1,0 im). Gii h phng trnh 2 2

ln(1 2 ) ln(1 2 ) 2 2

6 3 8 0

x y x y

x xy y ( ,x y ).

------------ Ht ------------

B GIO DC V O TO

THI TH




Cu 1 (2,0 im). Cho hm s 3 21 2

6( 1)3 3

y x mx m x c th l ( ),mC vi m l tham s thc.

a) Kho st s bin thin v v th hm s khi 1.m

b) Tm m trn ( )mC c hai im phn bit M 1 1( , )x y v N 2 2( , )x y sao cho tip tuyn ca ( )mC ti mi im

vung gc vi ng thng 3 6 0x y v 1 2 2 3.x x

Cu 2 (1,0 im). Gii phng trnh (1 sin ) tan cos2 cos sin 2 .x x x x x


2 2 2 3

22 2

4 ( 1 1)( 3 2)

1( 1) 2 1

x x x y y

xx y

y

( yx, ).


2

2

4

3cot 1.

sin

x xI dx

x

Cu 5 (1,0 im). Cho hnh chp S.ABCD c y ABCD l hnh ch nht BC 2AB. Hnh chiu vung gc ca S trn

mt phng (ABCD) l im H thuc cnh BD sao cho HD 3HB. Gc gia ng thng SB v mt phng (ABCD)

bng 060 . Bit khong cch t A n mt phng (SBD) bng

4.

5

a Tnh th tch hnh chp S.ABCD v tnh khong cch

gia hai ng thng SB v AD theo .a

Cu 6 (1,0 im). Cho cc s thc dng , ,a b c tha mn 2( )(b ) 4 .a b c c Tm gi tr nh nht ca biu thc

2

2

4 4 2 7 3.

a b ab c abP

b c a c c c

PHN RING (3,0 im): Th sinh ch c lm mt trong hai phn (phn A hoc phn B) A. Theo chng trnh Chun Cu 7.a (1,0 im). Trong mt phng vi h ta Oxy, cho hnh vung tm I. Gi M, N l lt l trung im ca AI

v BC. Bit N (6;3), phng trnh ng ng thng AC l 7 0x y v im P (4;7) nm trn tia DM. Tm ta

im D.

Cu 8.a (1,0 im). Trong khng gian vi h ta Oxyz, cho im M (1;2;3) v hai mt phng (P): 1 0,x y

(Q): 2 0.x z Tm ta im A thuc mt phng (P) sao cho MA song song vi mt phng (Q) v MA 3.

Cu 9.a (1,0 im). Xc nh x trong khai trin nh thc Niu-tn ca 2

2 ,2

n

x

x bit rng tng ca s hng th hai

v s hng th ba bng 135 v 2 1 22.n n nn n nC C C


Cu 7.b (1,0 im). Trong mt phng vi h ta Oxy, cho ng trn (C): 2 2 4 2 4 0x y x y v ng

thng : 2 0.d x y Gi I l tm ca (C), M l im thuc .d Qua M k cc tip tuyn MA, MB v ct tuyn MCD

n (C) (A, B l cc tip im v C nm gia M, D). Tm ta im M, bit 060 ,CED MC 4 v E l giao im ca AB vi IM.

Cu 8.b (1,0 im). Trong khng gian vi h ta Oxyz, cho ng thng 2 3 3

:4 2 1

x y zd v mt phng

( ) : 2 5 0.P x y z Vit phng trnh ng thng nm trong mt phng (P), song song vi ng thng d

v cch d mt khong bng 14.

Cu 9.b (1,0 im). Gii phng trnh 3 2 3 2 38 ( 3 6)4 2( 3 6) 0.x xx x x x

------------ Ht ------------

B GIO DC V O TO

THI TH





.2

xy

x

a) Kho st s bin thin v v th (C) ca hm s cho. b) Tm hai s thc m v n ng thng y mx n ct th (C) ti hai im phn bit A, B sao cho

26AB v AB vung gc vi ng thng : 2 2 1 0.x y

Cu 2 (1,0 im). Gii phng trnh sin 2 1

2sin tan .sin cos 2

xx x

x x

Cu 3 (1,0 im). Gii h phng trnh 2 2 2

1 1 8

( 2) ( 3) ( 5)

1 3 3 6

x y x y

x y x y

( ,x y ).

Cu 4 (1,0 im). Tnh din tch hnh phng gii hn bi cc ng 2 4 3y x x v 3.y x

Cu 5 (1,0 im). Cho hnh chp S.ABCD c y ABCD l hnh thang vung ti A v B, AD 2 AB 2 BC. Hnh chiu vung gc ca nh S ln mt phng (ABCD) l trung im H ca cnh AD. Bit gc hp bi hai mt phng

(SCD) v (ABCD) bng 060 ; khong cch t im B n mt phng (SCD) bng

6.

4

a Tnh th tch khi chp

S.ABCD v khong cch t D n mt phng (SAC) theo .a

Cu 6 (1,0 im). Cho , ,x y z l cc s thc thuc on [0;1] v tha mn iu kin 8 8 8 10.x y z Tm gi tr nh

nht ca biu thc 3 3 3 3( .2 .2 .2 ).x y zP x y z x y z


Cu 7.a (1,0 im). Trong mt phng vi h ta Oxy, cho hnh ch nht ABCD c din tch bng ,22 ng thng

AB c phng trnh ,0143 yx ng thng BD c phng trnh .032 yx Tm ta cc nh ca hnh

ch nht ABCD.


:1 2 2

x y zd v mt phng

( ) : 2 2 3 0.P x y z Tm ta giao im ca d v (P). Vit phng trnh mt phng cha ng thng d v

vung gc vi mt phng (P).

Cu 9.a (1,0 im). Tm s phc z tha mn cc iu kin 5 3z i iz v 2

izz

l s o.

B. Theo chng trnh Nng cao Cu 7.b (1,0 im). Trong mt phng vi h ta Oxy, cho tam gic ABC c bn knh ng trn ngoi tip tam gic

5 5,R cnh AB v ng trung tuyn AI ln lt c phng trnh l 0x y v 2 6 0.x y Tm ta cc

nh B v C ca tam gic, bit im M ( 3;9) nm trn ng thng AC.

Cu 8.b (1,0 im). Trong khng gian vi h ta Oxyz, cho hai ng thng 11 1 1

:1 2 2

x y zd v 2d l

giao tuyn ca hai mt phng (P): 2 1 0,x y (Q): 2 2 5 0.x y z Tm ta im I l giao im ca 1d v

2.d Vit phng trnh ng thng i qua A (2;3;1), ng thi ct 1,d 2d ln lt ti B v C sao cho tam gic

IBC cn ti I.

Cu 9.b (1,0 im). Vi s t nhin 2,n gi na l h s ca x trong khai trin nh thc Niu-tn ca (5 ) .nx

Tm gi tr ca n biu thc 2 3 4

2 3 4

5 5 5 5...

n

n

Aa a a a

c gi tr bng 48.

------------ Ht ------------

B GIO DC V O TO

THI TH




Cu 1 (2,0 im). Cho hm s 3 2 33( 1) 3 ( 2) 1y x m x m m x m (1), vi m l tham s thc.

a) Kho st s bin thin v v th hm s (1) khi 0.m b) Tm m th hm s (1) c hai im cc tr A v B sao cho tip tuyn ti im cc i A ca th ct trc

tung ti im M tha mn din tch tam OAM bng 2 (O l gc ta ).


11 (sin cos ) sin 2

12(1 cot ).

21 tan

4

x x x

x

x


21

2( 5 9) 1

x x

x x ( x ).


30

sin 2.

1 3sin 1 3sin

xI dx

x x

Cu 5 (1,0 im). Cho hnh hp ABCD.ABCD c y ABCD l hnh ch nht AB ,a BC 2 .a Hnh chiu vung

gc ca nh A ln mt phng (ABCD) l trung im H ca cnh BC. Bit rng gc gia hai mt phng (AABB) v

(ABCD) bng 045 . Tnh th tch khi hp ABCD.ABCD v khong cch gia hai ng thng CD v BB theo a.


3 3 2

2 2

4 2 3 2 1 7

4 (3 ) 2 3 4 7

y x x x y

y x y ( ,x y ).


Cu 7.a (1,0 im). Trong mt phng vi h ta Oxy, cho hai ng thng ,01:1 yxd 0322:2 yxd

v im N .2

3;

2

1 Vit phng trnh ng trn (C) tm I i qua im N, tip xc vi 1d v ct 2d ti hai im A, B

sao cho din tch tam gic IAB bng .8

3

Cu 8.a (1,0 im). Trong khng gian h ta Oxyz, cho mt phng (P): 0252 zyx

v cc im B ),0;1;1(

).0;5;1(C Xc nh ta im A thuc (P) sao cho tam gic ABC vung cn ti A.

Cu 9.a (1,0 im). Cho s phc z tha mn 22 2014 .z z z


Cu 7.b (1,0 im). Trong mt phng vi h ta Oxy, cho hnh vung ABCD tm I. Mt ng thng d i qua A

v ct cc cnh BC, CD ln lt ti M, N; ng thng IM ct BN ti E v ct CD ti F. Bit B (5; 4), 3 1

;2 2

E v

nh C nm trn ng thng : 6 2 11 0.x y Tm ta nh A.

Cu 8.b (1,0 im). Trong khng gian vi h ta Oxyz, cho mt cu (S): 2 2 2 2 4 3 0x y z x y v ng

thng 1

: .2 1 1

x y zd Vit phng trnh ng thng i qua A (1;0;0), vung gc vi ng thng d v tip

xc vi mt cu (S). Cu 9.b (1,0 im). Ty theo gi tr ca tham s ,m hy tm gi tr nh nht ca biu thc

2 2( 2) [4 2( 2) 1] .P x my x m y

------------ Ht ------------

B GIO DC V O TO

THI TH




Cu 1 (2,0 im). Cho hm s 4 22 3.y x x

a) Kho st s bin thin v v th (C) ca hm s.

b) Xc nh gi tr 3m ng thng y m ct th (C) ti hai im phn bit M, N sao cho tam gic IMN

vung ti I, vi im I (0;3).

Cu 2 (1,0 im). Gii phng trnh 2 24sin 3 cos 2 3 2cos .

2 4

xx x

Cu 3 (1,0 im). Gii phng trnh 2 22 9 3 3 7 1 3 2 0x x x x x (x ).


0

4

cos sin.

sin 2 2(sin cos ) 2

x xI dx

x x x

Cu 5 (1,0 im). Cho hnh chp S.ABCD c y ABCD l hnh vung cnh .a Gi M v N ln lt l trung im ca

cc cnh AB v AD; H l giao im ca CN vi DM. Bit SH vung gc vi mt phng (ABCD) v SH 3.a Tnh

th tch khi chp S.ABCD v tnh khong cch gia hai ng thng DM v SB theo .a

Cu 9 (1,0 im). Cho ba s thc dng , ,x y z thay i v tha mn iu kin 2 2 2 6.x y z xy yz zx Tm

gi tr nh nht ca biu thc 3 3 3

2 2 2

549ln( ) .

6

x y zP x y z

y z x xy yz zx


Cu 7.a (1,0 im). Trong mt phng vi h ta Oxy, cho hai ng thng vung gc 1 : 4 3 14 0,d x y

2 : 3 4 13 0d x y v im M ( 2;2). Vit phng trnh ng trn ( )C tip xc vi 1d ti M v ct 2d theo dy

cung AB 8.


:2 1 3

x y zd v mt phng

( ) : 2 2 3 0.P x y z Gi I l giao im ca d v (P). Tm ta im M thuc d sao cho hai im I, M v hnh

chiu ca M trn (P) l ba nh ca mt tam gic c din tch bng 13

.2

Cu 9.a (1,0 im). Cho s phc z tha mn 33 (2 ) (2 ).z z i i Tm phn o ca s phc .z


Cu 7 (1,0 im). Trong mt phng vi h ta Oxy, cho hnh ch nht ABCD im H (1;2) l hnh chiu vung gc

ca A ln BD. Bit M9

;32

l trung im ca cnh BC v phng trnh ng trung tuyn k t A ca tam gic ADH

l 4 4 0.x y Vit phng trnh ng thng BC.

Cu 8.b (1,0 im). Trong khng gian vi h ta Oxyz, cho ng thng 1 3 2

:1 1 1

x y zd v hai mt

phng (P): 2 3 0,x y z (Q): 2 2 1 0.x y z Tm ta giao im A ca d v (P). Vit phng trnh mt

cu (S) tm A sao cho (S) ct (Q) theo thit din l hnh trn c din tch bng 20 . Cu 9.b (1,0 im). C chn tm th nh s t 1 n 10. Chn ngu nhin ra hai tm th. Tnh xc sut tch ca

hai s trn hai tm th l mt s chn.

------------ Ht ------------

B GIO DC V O TO

THI TH




Cu 1 (2,0 im). Cho hm s 2 1

2

x my

x (1), vi m l tham s thc.

a) Kho st s bin thin v v th ca hm s (1) khi 2.m b) Vit phng trnh tip tuyn ca th hm s (1) ti giao im ca th hm s (1) vi trc tung, bit

khong cch t gc ta O n tip tuyn bng 2

.5

Cu 2 (1,0 im). Gii phng trnh 1 sin (1 sin )sin 2 cos2 .x x x x


3

2 2 2 2

2 2 1 3 1

5 2 12 7 19 5

y x x x y

x y x x y y ( ,x y ).


4

3

2

ln.

( 1)

x xI dx

x

Cu 5 (1,0 im). Cho lng tr ng ABC.ABC c y ABC l tam gic vung ti B, BC ,a AC 2 .a Gi M l

trung im ca cnh AC v N l im trn cnh BC sao cho CN 2BN. Gc to bi hai mt phng (CMN) v (ABC)

l vi 2

cos .4

Tnh th tch khi chp B.ABNM v khong cch gia hai ng thng BM v CN.

Cu 6 (1,0 im). Cho , ,a b c l cc s thc thuc on [1;9] v .x y z Tm gi tr nh nht ca biu thc

.3

x y zP

x y y z z x

PHN RING (3,0 im): Th sinh ch c lm mt trong hai phn (phn A hoc phn B) A. Theo chng trnh Chun Cu 7.a (1,0 im). Trong mt phng vi h ta Oxy, cho hnh vung ABCD vi M, N ln lt l trung im ca AB

v BC. Gi H l chn ng cao k t B xung CM. Bit N5

1; ,2

H ( 1;0) v im D nm trn ng thng

: 4 0.x y Tm ta cc nh ca hnh vung.

Cu 8.a (1,0 im). Trong khng gian vi h ta Oxyz, cho im A (3; 1;2) v ng thng d l giao tuyn ca

hai mt phng (P): 2 2 3 0,x y z (Q): 2 2 3 17 0.x y z Vit phng trnh mt phng i qua A v vung

gc vi .d Tm ta im i xng ca A qua .d

Cu 9.a (1,0 im). Cho n l s nguyn dng tha mn 2 2114 14 .n

n nA C n Tm s hng cha 5x trong khai

trin nh thc Niu-tn ca 2 1

,14

n

nx

x 0.x


Cu 7.b (1,0 im). Trong mt phng vi h ta Oxy, cho trn (C): 2 2 4 2 0x y x y v ng thng

: 2 0.x y Gi I l tm ca (C), M l im thuc v A (c tung m) thuc (C). Mt ct tuyn i qua M ct

(C) ti hai im B, C (im B nm gia M v C). Tm ta cc im A v M, bit din tch tam gic MAC bng ba ln

din tch tam gic MBC v ng phn gic trong ca gc A c phng trnh l 3 4 0.x y

Cu 8.b (1,0 im). Trong khng gian vi h ta Oxyz, cho mt phng (P): 6 31 0x y z v hai ng thng

1

2 1 5: ,

1 1 1

x y zd 2

2 3 2: .

3 3 1

x y zd Tm ta cc im A thuc 1,d B thuc 2d v C thuc (P)

sao cho AB BC t gi tr nh nht.

Cu 9.b (1,0 im). Tm s phc z tha mn 2 1 2z i z i v 1 10

.10z

------------ Ht ------------

B GIO DC V O TO

THI TH





x my

x (1), vi m l tham s thc.

a) Kho st s bin thin v v th hm s (1) khi 2.m b) Tm cc gi tr ca m ng thng 2 1y x ct th hm s (1) ti hai im phn bit A, B sao cho

2 2 2 34OA OB AB ( O l gc ta ).

Cu 2 (1,0 im). Gii phng trnh (sin 2 sin )(sin cos ) sin .x x x x x


11

1

x

x x x ( x ).


3 3

4 2

2

3.

5 6

x xI dx

x x

Cu 5 (1,0 im). Cho hnh chp S.ABC c y ABC l tam gic vung ti B, AB ,a BC 2.a Gi M, N ln lt

l trung im ca AC v BC. Bit rng SA vung gc vi mt phng (ABC) v gc gia hai mt phng (SBM) v (ABC)

bng 060 . Chng minh mt phng (SBM) vung gc vi mt phng (SAN) v tnh th tch khi chp S.ABC theo .a

Cu 6 (1,0 im). Cho cc s thc dng , ,a b c tha mn 1.a b c Tm gi tr nh nht ca biu thc

1 1 181 1 .

2P

a b a b c


Cu 7.a (1,0 im). Trong mt phng vi h ta Oxy, cho tam gic ABC cn ti A (4;6). im M (6;2) nm trn

cnh BC v trong tm G ca tam gic ABC nm trn ng thng : 2 2 0.x y Vit phng trnh cnh BC.


:1 3 8

x y z v cc im

(1;2;3),M C (0;0;1). Vit phng trnh mt phng (P) i qua C v ct cc tr Ox, Oy ln lt ti A, B sao cho (P)

song song vi v khong cch t M n (P) bng 8

,3

bit im B c tung khng nh hn 1

.2

Cu 9.a (1,0 im). Cho s phc z tha mn 1

2.1 2

zi

Tnh mun ca s phc ( 1)(2 )

.2

z iw

z i


Cu 7.b (1,0 im). Trong mt phng vi h ta Oxy, cho tam gic ABC c trc tm H ( 1; 7). Hai im D ( 5;7)

v E (1;10) ln lt l trung im ca AB v AC. Xc nh ta cc nh ca tam gic ABC, bit rng nh A c

honh l mt s nguyn.

Cu 8.b (1,0 im). Trong khng gian vi h ta Oxyz, cho im M (4;5; 1) v hai ng thng c phng trnh

1

4 3: ,

2 1 2

x y zd 2

3 3: .

2 1 1

x y zd Vit phng trnh ng thng i qua M v ct 1,d 2d ln lt

ti A v B sao cho MA 2MB.


2 1 1 1

3 2

2 4 2

3 3 2.9

log ( 1) log (2 1) log 2

x x y y

x y y ( ,x y ).

------------ Ht ------------

B GIO DC V O TO

THI TH




Cu 1 (2,0 im). Cho hm s 2 1

.2

xy

x


b) Tm m ng thng y x m ct th (C) ti hai im phn bit A, B sao cho tam gic OAB c bn

knh ng trn ngoi tip bng 2 2.

Cu 2 (1,0 im). Gii phng trnh .22

tantan1sin2cos2sin1

2cos2sin1 xxx

xx

xx

Cu 3 (1,0 im). Gii bt phng trnh 22 1 4 7 1 .x x x x


0

cos.

1 cos 2

x xI dx

x

Cu 5 (1,0 im). Cho hnh chp S.ABC c y l tam gic vung cn ti A, AB 2.a Tam gic SBC cn ti S v

nm trong mt phng vung gc vi mt phng (ABC). Gi M, N ln lt l trung im ca SB v AC. Bit gc gia

ng thng MN v mt phng (ABC) bng 030 . Tnh th tch khi chp S.ABC v khong cch gia hai ng thng

MN v BC theo .a

Cu 6 (1,0 im). Cho , ,a b c l cc s thc khng m v tha mn 1.a b c Tm gi tr nh nht ca biu thc

2 5.1 1

a bc b caP c

bc ca



Cu 7.a (1,0 im). Trong mt phng vi h ta Oxy, cho hnh vung ABCD c im M, N ln lt l trung im

ca BC v CD. Bit M (3;0) v phng trnh ng thng AN l 2 12 0.x y Vit phng trnh ng thng AD.


: .4 3 1

x y zd Vit phng

trnh mt cu c tm I (1;2; 3) v ct ng thng d ti hai im A, B sao cho AB 26.

Cu 9.a (1,0 im). Cho s phc z tha mn ( 1)(2 ) 3

.22

z i i

z i Tnh mun s phc (1 ) .w z z


Cu 7.b (1,0 im). Trong mt phng vi h ta Oxy, cho tam gic ABC c nh A (1;1); ng trung tuyn k t A

c phng trnh 3 2 5 0x y v ng thng i qua trung im ca cc cnh AB, AC c phng trnh

2 4 0.x y Tm ta cc nh B v C, bit din tch tam gic ABC bng 20 v B c honh dng.


:2 1 2

x y z v mt phng

( ) : 4 0.y z Vit phng trnh mt phng (P) vung gc vi ( ) v song song vi sao cho khong cch t

n (P) bng hai ln khong cch t gc ta O n (P).


2

2 2 2 2

4 2 2.16

log .log ( ) log log

x x y y

y x y x y ( yx, ).

------------ Ht ------------

B GIO DC V O TO

THI TH




Cu 1 (2,0 im). Cho hm s 3 23 1.y x x

a. Kho st s bin thin v v th (C) ca hm s.

b. Gi d l ng thng i qua im M ( 2;3) vi h s gc .k Tm k ng thng d ct th (C) ti ba

im phn bit sao cho cc tip tuyn ti ba giao im ct nhau to thnh tam gic vung.

Cu 2 (1,0 im). Gii phng trnh 3 sin 2cos cos 2 1 0.x x x

Cu 3 (1,0 im). Gii bt phng trnh 2 2 2(1 3) 2 (1 3) 2 3 2 2 2.x x x x x x


2

6

(sin 2 cos 1) (2 cos 1) ln.

sin ln

x x x x xI dx

x x x

Cu 5 (1,0 im). Cho hnh chp S.ABCD c y ABCD l hnh vung cnh .a Mt bn SAD l tam gic u v

2.SB a Gi E, F ln lt l trung im ca AD v AB; H l giao im ca FC v EB. Chng minh SE vung gc

vi EB, CH vung gc vi SB v tnh th tch khi chp C.SEB theo .a

Cu 6 (1,0 im). Cho , ,a b c l cc s thc khng m tha mn 1.a b c Tm gi tr nh nht ca biu thc

2 .P ab bc ca abc

PHN RING (3,0 im): Th sinh ch c chn lm mt trong hai phn (phn A hoc phn B)

A. Theo chng trnh chun

Cu 7.a (1,0 im). Trong mt phng vi h ta Oxy, cho hai ng thng : 2 0,d x y : 4 0x y ct

nhau ti M v cho im A (4;3). Tm ta cc im B, C ln lt thuc d v sao cho A l tm ng trn ngoi

tip tam gic MBC.

Cu 8.a (1,0 im). Trong khng gian vi h ta Oxyz, vit phng trnh ng thng i qua im A (3; 2; 4),

song song vi mt (P): 3 2 3 7 0x y z v ct ng thng 2 4 1

: .3 2 2

x y zd

Cu 9.a (1,0 im). Tm tp hp cc im biu din trong mt phng phc Oxy ca s phc ' (1 3) 2z i z bit

rng s phc tha mn 1 2.z


Cu 7.b (1,0 im). Trong mt phng vi h ta Oxy, cho hai ng thng 1 : 1 0,d x y 2 : 2 5 0d x y

v im M (1; 1). Gi A l giao im ca 1d v 2.d Vit phng trnh ng thng d i qua im M, ct hai ng

thng trn ln lt ti B v C sao cho tam gic tam gic ABC c BC 3AB.

Cu 8.b (1,0 im). Trong khng gian vi h ta Oxyz, cho hai ng thng 11 2 3

:2 2 3

x y z v

2

1 1 2: .

1 2 1

x y z Vit phng trnh mt cu c tm thuc 1, bn knh bng 5 ng thi ct 2 to thnh

mt dy cung c di ln nht.

Cu 9.b (1,0 im). Trong khai trin nh thc Niu-tn 1

2 ,

n

x h s ca s hng cha

2

1

x gp i h s ca s

hng th hai. Tm h s ca s hng cha 4

1

x v tnh tng h s ca tt c cc s hng khai trin.

----------Ht----------

B GIO DC V O TO

THI TH





.2 1

xy

x


b) Gi A l im biu din s phc 1z i v B l im biu din s phc 1 .w i Tm m ng thng :d y x m ct th (C) ti hai im phn bit C, D sao cho ABCD l hnh bnh hnh.

Cu 2 (1,0 im). Gii phng trnh 4 4sin cos 1

(tan cot 1).sin 2 2

x xx x

x


31

1 2( 3 4)

x x

x x ( x ).


2

0

.3cos sin cos 1

dxI

x x x

Cu 5 (1,0 im). Cho hnh chp S.ABCD c y ABCD l hnh thoi cnh 2 ,a 060 .BAD Hnh chiu vung gc ca nh S ln mt phng (ABCD) l trng tm H ca tam gic ABD v tam gic SAC vung ti nh S. Tnh theo a th tch khi chp S.ABCD v tnh c sin gc gia hai mt phng (SAC) v (SBC).

Cu 6 (1,0 im). Cho , ,a b c l cc s thc khng m tha mn 2 2 2 1.a b c Tm gi tr ln nht ca biu thc

2 2.

1 1 1 2

a c b c a b cP

bc ca abc


Cu 7.a (1,0 im). Trong mt phng vi h ta Oxy, cho hnh thoi ABCD c tm I (1; 2) v AC 2 BD. Bit

im M ( 5; 4) thuc ng thng AB v im N ( 5;16) thuc ng thng CD. Tm ta im B, bit honh

im B l mt s nguyn.


:2 1 2

x y zd v im

(2;1; 3).A Tm ta im A i xng vi A qua .d Vit phng trnh ng thng i qua A song song vi mt

phng (Oyz) ng thi ct .d Cu 9.a (1,0 im). Trong mt phng ta Oxy, tm tp hp im biu din cc s phc z tha mn

3 3 10.z z


Cu 7.b (1,0 im). Trong mt phng vi h ta Oxy, cho ng trn (C): 2 2 2 4 5 0x y x y c tm l I.

Vit phng trnh ng thng cch gc ta O mt khong bng 5 v ct ng trn (C) ti hai im phn

bit A, B sao cho tam gic IAB c din tch ln nht.

Cu 8.b (1,0 im). Trong khng gian vi h ta Oxyz, cho hai ng thng cho nhau 1

9

2

3

1

7:1

zyxd

v .3

1

2

1

7

3:2

zyxd Gi l ng vung gc chung ca 1d v 2.d Tnh khong cch t im A (1; 2;1)

n ng thng .


2 2 2

2 22

1log (2 ) log log ( 2)

2

4 4 2 16 2 12

x y y x

x x y x

( ,x y ).

----------Ht----------

B GIO DC V O TO

THI TH




Cu 1 (2,0 im). Cho hm s 3 2( 2) ( 1) 2 1y x m x m x m (1), vi m l tham s thc.

a) Kho st s bin thin v v th hm s (1) khi 1.m b) Tm m tip tuyn ca th hm s (1) ti im c honh 1x v ng thng : 2 1 0d x y to

vi nhau mt gc 030 .

Cu 2 (1,0 im). Gii phng trnh 1 2sin 1 sin

.1 2sin 3 cos

x x

x x

Cu 3 (1,0 im). Gii bt phng trnh 2 22(4 6) 2 1 2 3.x x x x x


0

( cos )sin.

1 cos 2

x x xI dx

x

Cu 5 (1,0 im). Cho hnh lng tr u tam gic ABC.ABC c cnh y bng .a Khong cch gia hai ng

thng BC v AC bng 15

.10

a Tnh th tch khi lng tr ABC.ABC v gc gia hai mt phng (ABC) v (CAB).

Cu 6 (1,0 im). Cho cc s thc dng tha mn 3.a b c Tm gi tr ln nht ca biu thc

2 2 2

( ) ( ) ( ).

a b c b c a c a bP a b c

a bc b ca c ab


Cu 7.a (1,0 im). Trong mt phng vi h ta Oxy, cho tam gic gic ABC vung ti A c nh B )1;1( v ng

thng AC c phng trnh .03234 yx Trn tia BC ly im M sao cho BC.BM .75 Tm ta nh C bit

bn knh ng trn ngoi tip tam gic AMC bng 5 5. Cu 8.a (1,0 im). Trong khng gian vi h ta Oxyz, cho mt phng (P): 2 2 9 0x y z v ng thng

1 3 3: .

1 2 1

x y zd Vit phng trnh mt cu c tm thuc ng thng , bn knh bng 2 v tip xc vi

mt phng (P).

Cu 9.a (1,0 im). Cho s phc z tha mn 2(1 2 ) . 4 20.i z z i Tnh mun ca s phc

1.

z


Cu 7.b (1,0 im). Trong mt phng vi h ta Oxy, cho hnh ch nht ABCD c nh B thuc ng thng

1 : 2 2 0d x y v nh C thuc ng thng 2 : 5 0.d x y Gi H l hnh chiu vung gc ca B xung

ng cho AC, bit M9 2

; ,5 5

K (9;2) ln lt l trung im ca AH v CD. Tm ta cc nh ca hnh ch nht

ABCD bit honh im C ln hn 4.

Cu 8.b (1,0 im). Trong khng gian vi h ta Oxyz, cho mt cu (S): 2 2 2 6 2 2 2 0x y z x y z v hai

ng thng 12 2 3

: ,2 1 1

x y zd 2

1 1 1: .

1 2 1

x y zd Vit phng trnh ng thng vung gc vi

hai ng thng 1,d 2d v ct (S) ti hai im A, B sao cho AB 6.


2

2 3

2 3

log (6 3 2 ) log ( 4 4) 4

log (5 ) log ( 1) 1

x y

x y

x y xy x x

y x ( yx, ).

------------ Ht ------------

B GIO DC V O TO

THI TH




Cu 1 (2,0 m). Cho hm s 3 1

.1

xy

x


b) Tm k ng thng ( 1) 2y k x k ct th (C) ti hai im phn bit A, B sao cho khong cch t

A v B n trc honh bng nhau.

Cu 2 (1,0 im). Gii phng trnh 2cos 2 1

1 tan cos sin 2 .cot 1 2

xx x x

x

Cu 3 (1,0 im). Gii h phng trnh 3 3

1 1

( 4 )(2 4) 36

x yx y

x y x y

( ,x y ).


1

4 2

0

.3 4

dxI

x x

Cu 5 (1,0 im). Cho hnh chp S.ABC c y ABC l tam gic vung ti A, AB ,a AC 2 .a Gi M v N ln lt

l trung im ca AB v BC. Hnh chiu vung gc ca S trn mt phng (ABC) l trung im H ca on MN; gc

gia ng thng SB v mt phng (ABC) bng 045 . Tnh th tch ca khi chp S.ABC v khong cch gia hai

ng thng MN v SC theo .a

Cu 6 (1,0 im). Cho , ,x y z l cc s thc dng tha mn 2 2 2 .x y z xyz Tm gi tr ln nht ca biu thc

2 2 2

1 1 1.

3 3 3

x y zP

x yz x y zx y z xy z xy yz zx

PHN RING (3,0 im): Th sinh ch c lm mt trong hai phn (phn A hoc phn B) A. Theo chng trnh Chun Cu 7.a (1,0 im). Trong mt phng vi h ta Oxy, cho hnh ch nht ABCD c phng trnh hai cnh BC v AD

ln lt l 4 3 3 0x y v 4 3 17 0;x y giao im ca hai ng cho AC v BD thuc ng thng

1 0.x y Vit phng trnh cnh AB, bit rng BC 3CD.

Cu 8.a (1,0 im). Trong khng gian vi h ta Oxyz, cho cc im A ( ;0;0),a B (0; ;0),b C (0;0;1), trong

,a b dng v mt phng (P): 4 3 0.x y z Xc nh a v b bit mt phng (ABC) vung gc vi mt phng

(P) v khong cch t im O n mt phng (ABC) bng 1

.3

Cu 9.a (1,0 im). Tm s phc z tha mn 2 2z v ( )( 2)z i z c phn o bng 2.


Cu 7.b (1,0 im). Trong mt phng vi h ta Oxy, cho hnh thoi ABCD c nh A (5;5); ng thng i qua

trung im ca cc cnh BC v CD c phng trnh 14 0.x y Tm ta cc nh B, C, D ca hnh thoi, bit

im E (0; 4) nm trn ng thng i qua D v vung gc vi AB.

Cu 8.b (1,0 im). Trong khng gian vi h ta Oxyz, cho mt cu (S): 2 2 2 4 2 6 10 0x y z x y z v

ng thng 1 2

: .2 1 3

x y zd Tm ta im M thuc mt cu (S) sao cho khong cch t M n ng thng

d l ln nht.

Cu 9.b (1,0 im). Xc nh x trong khai trin nh thc Niu-tn ca 2

2log

log 34 82 ,

nx

xxx bit rng s hng th

3 trong khai trin bng 45 v n l s nguyn dng tha mn 1 2 2 2 1 20

2 1 2 1 2 1 2 1... 2 .n n n n

n n n nC C C C

------------ Ht ------------

B GIO DC V O TO

THI TH




Cu 1 (2,0 im). Cho hm s 2 1

.1

xy

x

a) Kho st s bin thin v v thi (C) ca hm s.

b) Cho im E (1;0). Tm im M thuc (C) sao cho tip tuyn ca (C) tai M ct ng tim cn ngang ti F v

tam gic EFM vung ti F.

Cu 2 (1,0 im). Gii phng trnh ).1cos2sin3(2

1cos)12cot(tan 3 xxxxx

Cu 3 (1,0 im). Gii bt phng trnh 22 2(5 17 20) 3 .x x x x


2

0

sin.

2cos 5sin cos

xI dx

x x x

Cu 5 (1,0 im). Cho hnh chp S.ABCD c y ABCD l hnh ch nht, AB 2 ,a AD .a Trn cnh AB ly im

M sao cho AM ,4

AB cnh AC ct MD ti H. Bit SH vung gc vi mt phng (ABCD) v SH .a Tnh th tch

khi chp S.HCD v khong cch gia hai ng thng SD v AC theo .a

Cu 6 (1,0 im). Cho cc s thc , ,x y z tha mn iu kin 2 2 2 1.x y z Tm gi tr nh nht ca biu thc

5 4 2 4 2( ).P x x y z

PHN RING (3,0 im): Th sinh ch c lm mt trong hai phn (phn A hoc phn B) A. Theo chng trnh Chun Cu 7.a (1,0 im). Trong mt phng vi h ta Oxy, cho tam gic ABC cn ti B c phng trnh cnh AB l

2 3 0.x y Bit rng M (1;0) l trung im ca AC v AC 2 5

5BC. Tm ta nh B.

Cu 8.a (1,0 im). Trong khng gian vi h ta Oxyz, cho im M (5; 1; 13) v hai ng thng c phng

trnh 1

5 1 13: ,

2 3 2

x y zd 2

7 3

: 1 2 .

8

x t

d y t

z

Vit phng trnh mt phng (P) song song vi 1,d 2d v

khong cch t M n mt phng (P) bng 308.

Cu 9.a (1,0 im). Cho n l s nguyn dng tha mn 2 3

2 14 1.

3n nC C n Tm h s ca s hng cha

9x trong khai

trin nh thc Niu-tn ca 2(1 3 ) .nx


Cu 7.b (1,0 im). Trong mt phng vi h ta Oxy, cho tam gic ABC c trc tm H ),5;5( phng trnh ng

thng cha cnh BC l .08yx Bit ng trn ngoi tip tam gic ABC i qua hai im M )3;7( v ).2;4(N

Tnh din tch tam gic ABC.

Cu 8.b (1,0 im). Trong khng gian vi h ta Oxyz, cho mt phng (P): 3 0x y z v ng thng

1 2: .

1 1 1

x y zd Gi A l giao im ca d v (P). Vit phng trnh ng thng i qua im E (2;1;1), ct

d v (P) ln lt ti M v N sao cho tam gic AMN cn ti M.

Cu 9.b (1,0 im). Gii h phng trnh 142log)13(log

222

2

5

2

5

12

yxyyx

xyxy

yx,( ).

------------ Ht ------------

B GIO DC V O TO

THI TH




Cu 1 (2,0 im). Cho hm s 4 2 22y x mx m m (1), vi m l tham s thc.


b) Tm m th hm s (1) ct trc honh ti 4 im phn bit c honh 1,x 2 ,x 3 ,x 4x tha mn iu

kin 4 4 4 41 2 3 4 8.x x x x


(1 sin 2 )sin14

sin .1 cot 2

x x

xx

Cu 3 (1,0 im). Gii h phng trnh 2 2 2 2 2 2

(3 ) 4

( )( ) 5 2 1

x x y

x y xy x y x y ( yx, ).

Cu 4 (1,0 im). Tnh din tch hnh phng gii hn bi cc ng 2

1

1y

x v

2

.2

xy

Cu 5 (1,0 im). Cho hnh chp S.ABCD c y ABCD l hnh vung cnh ;a mt bn SAB l tam gic cn ti S v

nm trong mt phng vung gc vi y. Gc gia ng thng SC v mt phng (ABCD) bng 060 . Gi M v H ln

lt l trung im ca cc cnh SD v AB; I l giao im ca DH v AC. Tnh th tch khi t din MAID v tnh bn knh mt cu ngoi tip hnh chp S.ABCD theo .a

Cu 6 (1,0 im). Cho cc s thc khng m , ,a b c tha mn 2 2 2 2.a b c Tm gi tr ln nht ca biu thc 2

2

1.

1 1 9

a b c bcP

a bc a a b c

PHN RING (3,0 im): Th sinh ch c lm mt trong hai phn (phn A hoc phn B) A. Theo chng trnh Chun Cu 7 (1,0 im). Trong mt phng vi h ta Oxy, cho hnh thang cn ABCD c hai ng cho vung gc vi

nhau ti H. Gi M l im trn cnh AB sao cho AB 3AM, N l trung im ca HC. Gi s B ( 1; 3), ng thng

HM i qua im E (2; 3) v ng thng DN c phng trnh 2 5 0.x y Tm ta cc nh A v D.


:2 1 3

x y zd v im

(2; 5; 6).A Gi B l im thuc d v B c honh ln hn 1 sao cho AB 35. Vit phng trnh mt cu (S)

c ng knh l on AB.

Cu 9.a (1,0 im). Tm s phc ,z bit 2 5z i v im biu din s phc z thuc ng thng

3 1 0.x y


Cu 7.b (1,0 im). Trong mt phng vi h ta Oxy, cho ng trn (C): 2 2 4 2 0x y x y v ng thng

: 2 12 0.x y T mt im M nm trn ng thng k hai tip tuyn MA, MB n ng trn (C) (A v B l

hai tip im). Vit phng trnh ng trn ngoi tip tam gic MAB, bit rng AB 15.

Cu 8.b (1,0 im). Trong khng gian vi h ta Oxyz, cho im A (1;1;1), ng thng 1 3

:2 1 1

x y zd v

mt phng (P): 2 4 0.x y z Vit phng trnh ng thng qua A, vung gc vi d v ct (P) ti im B

sao cho AB 6. Cu 9.b (1,0 im). Tm s nguyn dng n tha mn

0 1 21 1 1 1 15701... .3 4 5 1 220

n

n n n nC C C Cn

------------ Ht ------------

B GIO DC V O TO

THI TH




Cu 1 (2,0 im). Cho hm s 2 3

.2

xy

x

a) Kho st s bin thin v v th (C) ca hm s. b) Vit phng trnh cc tip tuyn ca th (C), bit tip tuyn ct cc ng tim cn ln lt ti A v B

sao cho AB 8.

Cu 2 (1,0 im). Gii phng trnh .4cos1

sin42cot1

2

x

xx

Cu 3 (1,0 im). Gii phng trnh 2 21 1(5 6)

5 7 1x x

x x ( x ).


4

3

.2 6.2 5x x

dxI

Cu 5 (1,0 im). Cho hnh chp S.ABCD c y ABCD l hnh vung tm O cnh a; SA vung gc vi mt phng

(ABCD) v 060 .BSD Gi M l trung im ca cnh BC. Tnh (theo a) th tch khi chp S.CDOM v tnh gc gia hai mt phng (SBC) v (SBD).

Cu 6 (1,0 im). Cho , ,x y z l cc s thc khng m v tha mn iu kin , , 1x y z v 2 2 2 12.x y z Tm

gi tr nh nht ca biu thc

2 22log loglog 1.

3

y xzx zy

P x y zx y y z z x

PHN RING (3,0 im): Th sinh ch c lm mt trong hai phn (phn A hoc phn B) A. Theo chng trnh Chun Cu 7.a (1,0 im). Trong mt phng vi h ta Oxy, cho tam gic ABC vung ti A, ng thng AB v ng

thng cha trung tuyn AM ca tam gic ln lt c phng trnh l 4 3 1 0x y v 7 8 0.x y Tm ta

cc nh ca tam gic ABC, bit im E (10;3) thuc ng thng BC.

Cu 8.a (1,0 im). Trong khng gian vi h ta Oxyz, cho mt cu (S): 2 2 2 4 4 8 1 0x y z x y z v

ng thng 2 2 3

: .1 3 2

x y zd Chng minh rng d v (S) ct nhau ti hai im phn bit. Vit phng trnh

mt cu (T) c cng tm vi (S) v tip xc vi .d Cu 9.a (1,0 im). Cho n l s nguyn dng tha mn 2 11 5.

n

n nA C Tm h s ca 5x trong khai trin biu thc

2 2(1 2 ) (1 3 ) .n nP x x x x


Cu 7.b (1,0 im). Trong mt phng vi h ta Oxy, cho hnh vung ABCD c din tch bng 16. Gi M (2;3) l

im nm trn ng cho AC, E v F ln lt l hnh chiu vung gc ca M ln AD v DC. Bit ng thng EF c

phng trnh 3 6 0,x y im I7

;22

l trung im ca on thng BF v nh C c tung nh hn 1. Tm

ta cc nh ca hnh vung ABCD.

Cu 8.b (1,0 im). Trong khng gian vi h ta Oxyz, cho mt phng (P): 2 2 0x y z v ng thng

1 2: .

2 1 1

x y zd Gi A l im trn d sao cho khong cch t A n (P) bng 1 v B l im thuc (P) sao

cho AB vung gc vi d v di AB nh nht. Tm ta cc im A v B. Cu 9.b (1,0 im). Cho s phc z tha mn cc iu kin: z l s thun o v z l mt nghim ca phng trnh

3 22(1 ) 4(1 ) 8 0.z i z i z i Tnh mun ca s phc .z

--------------- Ht ----------------

B GIO DC V O TO

THI TH




Cu 1 (2,0 im). Cho hm s .2323 xxy


b) Tm tt c cc im M nm trn ng thng 2y sao cho t M ta k c ba tip tuyn vi th (C).


6 62(1 cos sin ) 3sin cos0.

2 2sin

x x x x

x

Cu 3 (1,0 im). Tm m h phng trnh 2 5

5 1

xy y x y

x y m c nghim thc.

Cu 4 (1,0 im). Cho hnh H gii hn bi cc ng 5

,1 3 2

xy

x 0,y 1x v 3.x Tnh th tch ca

khi trn xoay to thnh khi quay hnh H xung quanh trc Ox. Cu 5 (1,0 im). Cho hnh chp S.ABC c y ABC l tam gic vung ti B v AB ;a hai mt phng (SAB) v

(SAC) cng vung gc vi mt phng (ABC). Gc gia ng thng SC v mt phng (SAB) bng 060 . Gi M l trung

im ca cnh AC. Bit khong cch gia hai ng thng SM v AB bng 6

.2

a Tnh th tch ca khi chp S.ABC.

Cu 6 (1,0 im). Cho cc s thc dng , ,a b c tha mn 2 2 2 3.a b c Tm gi tr ln nht ca biu thc

.4 4 4

ab bc caP

ab bc ca


Cu 7.a (1,0 im). Trong mt phng vi h ta Oxy, cho ng trn (C): 2 2 2 4 4 0x y x y v ng

thng :3 4 19 0.x y T mt im M nm trn ng thng k hai tip tuyn MA, MB n ng trn (C) (A

v B l hai tip im) sao cho 060 .AMB Vit phng trnh ng trn ngoi tip tam gic MAB.

Cu 8.a (1,0 im). Trong khng gian vi ta Oxyz, cho im M (1;2; 1) v ng thng 2 4

: .1 1 1

x y zd

Vit phng trnh mt phng (P) i qua im O, song song vi ng thng d ng thi cch im M mt khong

bng 2.

Cu 9.a (1,0 im). Tm s phc z tha mn iu kin 11 2 3

2 0.i

zz


Cu 7.b (1,0 im). Trong mt phng vi h ta Oxy, cho hnh vung ABCD c im A (1;4), im M( (2;3) nm

trn ng cho AC. Gi E v F ln lt l hnh chiu vung gc ca M trn AD v CD. Bit ng thng EF v CE

ln lt c phng trnh 2 5 0x y v 2 3 11 0.x y Tm ta cc nh ca hnh vung ABCD.

Cu 8.b (1,0 im). Trong khng gian vi h ta Oxyz, cho mt cu (S): 2 2 2 2 4 6 5 0x y z x y z v

im A (0;4;1). Vit phng trnh ng thng tip xc vi mt cu (S) ti im A v to vi trc Oy mt gc ,

bit rng 1

os .3 10

c

Cu 9.b (1,0 im). Cho khai trin

1 22 2 2 2

0 1 2 2 ... ( 1) .2 2 2 2

n n n n

n n n

n n n n

nx nx nx nxx C C x C x C x

(vi n l s nguyn dng). Bit trong khai trin 3 23 10n nC C v t s gia s hng th 3 v s hng th 4 bng 9.

Tm n v .x

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25 Đề Thi Thử Đại Học Môn Toán 2015 Có Lời Giải Và Đáp Án Chi Tiết

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Transcript of 25 Đề Thi Thử Đại Học Môn Toán 2015 Có Lời Giải Và Đáp Án Chi Tiết