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s 4

I. PHCu I m s .

1. Kh s n v v2. Tm m nh c 6 nghi

Cu II

1. Gi nh: (1)

2. Tm m nh sau c nghi :

(2)

Cu III

Cu IV .A1B1C1 c AB = a, AC = 2a, AA 1 v. G 1. Ch MA 1 v tnh

kho 1BM).Cu V Cho x, y, z l cc sII. PH

A nh ChuCu VI.a. , cho cc

v n tr

ph1. Cho . Tm gc gi m .2. Tm a N nh

Cu VII.a. nh:

B nh Nng cao.Cu VI.b. 1; 3; 2), B (3; 7; 18) v

m 2x y + z + 1 = 01. Vi trnh m vung gc v2. Tm t (P) sao cho MA + MB nh

Cu VII. b. nh:


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s 5

I. PH

Cu I Cho hm s

1. Kh n v v m s2. V i Av B. Gl giao hai ti . Tm v

Cu II

1. Gi nh: (1)

2. Gi nh : (2)

Cu III nh tch phn sau:

Cu IV Cho hnh chp t n b , m n h. Tm kh

Cu V Cho x, y, z l cc s m gi tr

II. PHA. T nh chu

Cu VI.a

1. Trong m , cho hnh ch ; 0) .

nh x 2y + 2 = 0, AB = 2AD. Tm to

2. Trong khng gian v , v g

trnh:

L nh m (P) ch ) v .Cu VII.a Tm m nh sau c 2 nghi

(3) nh nng cao

Cu VI.b 1. Trong m Oxy, cho hnh vung ABCD bi 2);P(2;0); Q(1;2) l y l nh cc chnh vung.2. Trong khng gian v , ) v ( ) g

trnh:

Vi nh ) v ( ).Cu VII.b Gi bi nh:(4)


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s 6I. PHCu 1 ):Cho hm s 3 3 ( 1 ) y x x

1) Kh n v v m s2) Ch x +1) + 2 lun c

(C) tbi P vung gc vCu 2 ):

1) Gi nh: 2 1 1 15 .3 7 .3 1 6 .3 9 0 x x x x (1) 2) Tm t nh sau c 2 nghi

(2)

Cu 3 ): Gi nh: (3)

Cu 4 ): Cho hnh chp S.ABCD c nh chbn c nh chp b b 2a . G

c n c3a

AK . Hy tnh kho

SK theo a.Cu 5 n: a + b + c =1. Tm gi tr

th .

II. PH nh chu

Cu 6a 1) Trong m 2y + 2 =0. Tm trn d hai AB = 2BC.2) Trong khng gian v nh: x 2 + y 2 + z 2 2x + 4y + 2z 3 = 0 v m y + 2z 14 = 0. Vi nh mph c n c bn knh b

Cu 7a m cc sT nh: trn tTm m

nh nng caoCu 6b

1) Trong m n (C): x 2 + y 2 6x + 5 = 0. Tm gc gi

ti 0.2) Trong khng gian v

(d1) : ; (d 2) : 3 ; ; 0 x t y t z

Ch 1) v (d 2) cho nhau. Vi nh m d1) v (d 2).

Cu 7b Cho s ln2. Tnh J = v tm


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s 7

I. PHCu I ( ):Cho hm s 3 22 ( 3) 4 y x mx m x (C m).

1) Kh n v v 1) c m s n khi m = 1. nh y = x + 4 v m cc gi trc m) t gicKBC c di 8 2 .

Cu II ( ):1) Gi nh: cos2 5 2(2 cos )(sin cos ) x x x x (1)

2) Gi nh: 3 3 3

2 2

8 27 18

4 6

x y y

x y x y(2)

Cu III ( ): Tnh tch phn: I =2 2

6

1sin sin

2 x x dx

Cu IV ( ):Cho hnh chp S.ABC c gc gi (ACB) b600

Cu V Tm cc gi tr nh sau c nghi2 21 1 1 19 ( 2)3 2 1 0 x xm m (3)

II. PH nh chu

Cu VIa (2 ):1) Trong m n (C) c ph nh

2 21 2 9 m mc duy nh t n (C)(B, C l hai ti2) Trong khng gian v

nh: 1 12 1 3

x y z . L nh m

d v kho lCu VIIa ( ):Cho ba s n abc = 1. Ch

3 3 34 4 43

(1 )(1 ) (1 )(1 ) (1 )(1 )a b c

b c c a a b(4)

trnh nng cao:Cu VIb ( ):

1) Trong m v c 3), B(3;2), tam gic ABC cdi 3

2; tr ABC n y 8 = 0.

Tm bn knh n n p ABC.2) Trong khng gian v giao tuyph 2y z + 1 = 0, (Q): x + 2y 2z 4 = 0 v m 2 + y 2 + z 2 + 4x 6y + m = 0. Tm m i MN = 8.

Cu VIIb Gi h nh :2 2

2 22 2log ( ) 1 log ( )

3 81 x xy y

x y xy(x, y R)


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s 8

I. PHCu I m s 4 2 2( ) 2( 2) 5 5 f x x m x m m (Cm)

1) Kh n thin v v m s2) Tm m m nh 1 tam gic vung cn.

Cu II: 1) Gi nh sau trn t 1 1

2 3 5 2 x x x(1)

2) Tm cc nghi ng trnh sau tho n 13

1 log 0 x :

sin .tan 2 3(sin 3 tan 2 ) 3 3 x x x x (2)

Cu III1

0

12 ln 1

1

x I x x dx

x

Cu IV: nh chp S.ABCD c nh thoi v 0120 A , BD = a>0. C n SA vung gc v 0. Mmc nh chp do m nh chp.

Cu V: n abc a c b . Hy tm gi trnh 2 2 2

2 2 31 1 1

Pa b c

(3)

II. PH NG nh chu

Cu VI.a:1) Trong m cho tam gic ABC cn, ctrnh 1 0 x y nh :2 2 0 x y

nh cc c n c2) Trong khng gian v vi nhM(1;1;1), c 1

2 1:

3 1 2 x y z

d v vung gc v

2 : 2 2 ; 5 ; 2d x t y t z t ( t R ).Cu VII.a: nh: 1 2 3 23 7 ... (2 1) 3 2 6480n n n nn n n nC C C C

B. nh nng caoCu VI.b:

1) Trong m 2 2

5 5 x y , Parabol2

( ) : 10P x y .Hy vi nh n c tm thu ( ) : 3 6 0 x y th nh Ox v ct tuy2) Trong khng gian v nhv 1 0 x y z

11 1

:2 1 1

x y zd v 2( ) : 1 ; 1;d x t y z t , v t R .

Cu VII.b: nh sau trn t2

4

2 2 1

1 6log ( )

2 2 ( ) x x x y a

y y b. (4)


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s 9

I. PHCho hm s 3 + (1 2m)x 2 + (2 m)x + m + 2 (m l tham s

1) Kh n v v m s2) Tm cc gi tr m shonh

1) Gi nh: 3 3 2 3 2cos3 cos sin 3 sin8

x x x x (1)

2) Gi nh:2

2

1 ( ) 4

( 1)( 2)

x y y x y

x y x y(x, y ) (2)

Tnh tch phn:5

3 2 1 4 1dx I

x x

Cho hnh h 32

a

v gc BAD = 60 0 . G N l AB.Ch ch khA.BDMN.

Cho x,y l cc s n 2+xy+y 2 3 .Ch2 24 3 3 3 4 3 3

II. PH nh chu

1) Trong m vd: x 4y 2 = 0, c nh

M(1; 1). Tm t2) Trong khng gian v ): 3x + 2y z + 4 = 0 v hai

K sao cho KI vung gc v ( ).

Gi nh:

nh nng cao

1) Trong m c c 1).Bi nh nh

m t .2) Trong khng gian v 3y + 11z = 0 v hai

1:1

x=

23y

=3

1z,

14x

=1y

=2

3z. Ch 1 v d 2

cho nhau. Vi nh n c 1 v d 2.

Gi nh: 14 2 2 2 1 2 1 2 0 .


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s 10

I. PH

Cu I . Cho hm s2

12

x

x y (C).

1) Kh n v v m s2) Ch ng d: y = x + m lun lun cbi m m i nh

Cu II 1) Gi nh: 9sinx + 6cosx 3sin2x + cos2x = 8

2) Gi nh: )3(log53loglog 242

222 x x x

Cu III Tm nguyn hm x x

dx I

53 cos.sin

Cu IV 1B1C1 c t

c n v m0

. Hnh chi n m(A1B1C1) thu 1C1. Tnh kho 1 vB1C1 theo a.

Cu V Cho ba s n: a 2009 + b 2009 + c 2009 = 3. Tm gitr P = a 4 + b 4 + c 4.

II. PH NG nh chu

Cu VIa 1) Trong m 1): 7 17 0 x y , (d 2):

5 0 x y . Vi nh 1), (d 2) mtam gic cn t 1), (d 2).2) Trong khng gian v nh hA O, B(3;0;0), D(0;2;0), A(0;0;1). Vi nh m

Cu VIIa . C bao nhiu s n c 4 ch khc 0 m trong ms hai ch

nh nng cao )Cu VIb

1) Trong m nhth c 1): x + y + 1 = 0, (d 2): x 2y + 2 = 0 l

2) Trong khng gian v 1), (d 2)

v 1):1 2

3 2 1 x y z

; (d 2) l giao tuy 1 0 x v (Q):

2 0 x y z . Vi nh 1) v c 2).

Cu VIIb Tm h 8 x khai tri 2 3 8(1 )P x x .


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s 11

I. PHCu I m s 1

1 x

y x

(C).

1) Kh n v v m s2) Tm trn tr m

Cu II 1) Gi nh: 2 2 2 2 2log ( 1) ( 5) log( 1) 5 0 x x x x 2) Tm nghi nh: 2 3cos sin 2 x cos x x tho n : 1 3 x

Cu III 1

2

0

ln( 1) I x x x dx

Cu IV nh l ABC l tam gic vung t AB = a , BC = b, AA = c ( 2 2 2c a b ). Tnh di nh lb vung gc v .

Cu V , , (0;1) x y z v 1 xy yz zx . Tm gi tr

bi 2 2 21 1 1 x y z

P x y z

II. PH nh chu :

Cu VI.a:1) Trong khng gian v nh:{ x t ; 1 2 y t ; 2 z t ( t R ) v m 2 2 3 0 x y z .Vitrnh tham s n n (P), c vung gc v

2) Trong m2 2

19 4

x y. Vi nh

Cu VII.a nh sau trn t 2 28

1

z w zw

z w

nh nng cao:Cu VI.b

1) Trong khng gian v 1), B(1;4;1), C(2;4;3),D(2;2;1). Tm t MA 2 + MB 2 + MC 2 + MD 2 2) Trong ml cc s C n n tr nh c

. Bi u vi c b m t

Cu VII.b Gi nh:2 1

2 1

2 2 3 1( , )

2 2 3 1

y

x

x x xx y R

y y y


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s 12

I. PHCu I m s 3 23 2 y x m x m (Cm).

1) Kh n v v m s m = 1 .2) Tm m m) v tr

Cu II 1) Gi nh: (sin 2 sin 4)cos 2 0

2sin 3

x x x

x

2) Gi nh: 3 18 1 2 2 1 x x

Cu III 2

30

sin(sin cos )

xdx I

x x

Cu IV (ABC), SC =a . Tnh gc gi

Cu V m m nh sau2 2 (2 )(2 ) x x x x m

II. PHng trnh chu

Cu VI.a 1) Trong m nhth B sao cho (OA+3OB) nh2) Trong khng gian v A(1;2;3) v B(3;4;1). Tm to

1 0 x y z

Cu VII.a m h 20 x trong khai tri 532

n

x x

,

bi t r 0 1 21 1 1 1... ( 1)2 3 1 13

n nn n n nC C C C n

nh nng cao:Cu VI.b

1) Trong m 2;4), C(1;4), D(3;5).Tm to ( ) : 3 5 0 x y sao cho hai tam gic MAB,MCD c di2) Trong khng gian v 1( ) nh

2 ; ; 4 x t y t z ; 2( ) l giao tuy ( ) : 3 0 x y v( ) : 4 4 3 12 0 x y z . Ch 1 2, cho nhau v vitrnh m 1 2,

Cu VII.b m s2 2(2 1) 4

2( ) x m x m m

y x m

. Ch m,

hm s kho m.


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s 13

I. PHCu I m s 3 1

2 4 x m y

m x m (C m) (m l tham s

1) Kh n v v m sx + m c

ngCu II

1) Gi trnh: s 4sin 2 1inx cosx x .

2) Tm m nh:2 2

2 2

2

4

x y x y

m x y x yc ba nghi

Cu III 1 3 20

1 I x x dx ; J =1

1( ln )

e x

x xe dx

x e x

Cu IV nh l ABCD.A'B'C'D' c n csao cho AM = x, (0 < x < a). M

kh 13

th

Cu V: Cho x, y l hai s 5 = 0. Tm gitr 4 1

4 x y.

II. PH nh ChuCu VI.a

1) Trong m h 1: 3 4 5 0 ; 2:4 3 5 0 . Vi nh n c tm n 6y 10 = 0 v ti 1, 2.2) Trong khng gian v nh chp A.OBC, trongthu(ABC) vung gc v tan 2OBC . Vi nh tham s

Cu VII.a Gi nh: 2 2(2 ) 7 4 0 z i z i trn t

nh Nng cao :Cu VI.b

1) Trong m 1(155; 48), M 2(159; 50),M3(163; 54), M 4(167; 58), M 5(171; 60). L nh

th cho nh2) Trong khng gian v mt hnh chtrnh m

Cu VII.b ( Ch 4 28 8 1 1a a , v 1 ; 1].


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s 14

I. PHCu I. m s 2 1

1 x

y x

(C)

1) Kh n v v m s2) Tm ccc nh

Cu II.

1) Tm m nh c nghi1

1 3

x y

x x y y m.

2) Gi nh: cos 23xcos2x cos 2x = 0.

Cu III. 220

( sin )cos I x x xdx.

Cu IV. n c nh vung ABCD c i l a, l

AM = x (0 m a). Trn n

v x. Tm gi tr 2 + y 2 = a 2.

Cu V. cc s n: 1 1 1 1 x y z

. Ch

1 1 11

2 2 2 z y z x y z x y z.

II. PH nh chu

Cu VI.a.

1) Trong m elip (E):2 2

14 1

x y . Tm

to

2) Trong khng gian v 2 + y 2 + z 2 2x + 2y + 4z

1 2

1 1: , :

2 1 1 1 1 1 x y z x y z . Vi nh ti

di S), bi 1 v 1.

Cu VII.a. Gi nh:2. 5. 90

5. 2. 80

x x y y

x x y y

A C

A C

nh nng caoCu VI.b.1) Trong m 2 = 8x. Gi

c x1, x2. Ch 1 + x 2 + 4.2) Trong khng gian vth c nh tham s 1 2 ; 1 ; 2 x t y t z t . M

nh

Cu VII.b. m c m s3

1( ) ln3

f x x

v gi bpt: .


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s 15

I. PH

Cu I m s3

3 y x x 1) Kh s n v v m s .2) Tm trn

Cu II1) Gi nh. : 3sin 2 2sin 2

sin2 .cos x x x x

2) Tm m nh sau c nghi ( 1) 4( 1)1

x x x x m

x

Cu III tch phn I= 22

sin 3

0

.sin .cos . xe x x dx.

Cu IV nh nn n AB = 2R.G n 2 ASB , 2 ASM . Tnh thdi theo R, v .

Cu V 2 2 2 1a b c . Ch 2(1 ) 0abc a b c ab ac bc II. PH

nh chuCu VI.a

1) Trong m n (C): (x 1) 2 + (y + 1) 2 = 25 v nh

phn bi2) Trong khng gian v 2).G hnh chi n m ph m t

Cu VIIa nh: 22 2log ( 7) log 12 4 0 x x x x nh nng cao

Cu VI.b 1) Trong m nh bnh hnh ABCD c diBi g mt D.2) Trong khng gian v ABC v

nh nh

1 2 3 3: 1 1 2 x y zd , 2 1 4 3: 1 2 1

x y zd .

L nh ABC v tnh di ABC .

Cu VII.b nh: 2008 2007 1 .


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s 16

I. PHCu I m s 2 4

1 x

y x

.

1) Kh n v v m s2) Tm trn (C) hai g th 3;0) v N(1; 1)

Cu II 1) Gi nh: 4cos 4x cos2x 1 3cos4 cos

2 4 x

x = 72

2) Gi nh: 3 x.2x = 3 x + 2x + 1

Cu III: K =2

0

1 sin.

1 cos x x e dx

x

Cu IV nh chp tam gic i c n b

bn h nh th nh c nh chp S.ABC.Cu V ba c h2 2 252 2 2

27a b c abc

II. PH NG: nh chu :

Cu VI.a 1) Trong m nh hai c 5x 2y + 6 = 0 v 4x + 7y 21 = 0. Vi nh ctr ng v2) Trong khng gian v m trn Ox

(d) : 1 21 2 2

x y z v m y 2z = 0

Cu VII.a m gi tr m s 2cos

sin (2cos sin ) x

x x xv

3.

nh nng cao:Cu VI.b

1) Trong m 3y 4 = 0 vn (C): x 2 + y 2 4y = 0. Tm M thu N thu

x

2) Trong khng gian v

2 4

3 2 2

x y z

1), B(7; 2;3). Tm trn (d) nh h t

B l nh

Cu VII.b 2 23 cos sin3 3

i . Tm cc s 3


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s 17

I. PHCu I: m s 2 1

1 x

y x

(C)

1) Kh n v v m s2) Tm m

OAB vung tCu II:

1) Gi nh:2cos . cos 1

2 1 sinsin cos

x x x

x x

2) Gi nh:2 2

2 2

3 ( )

1 1 4 ( )

x y xy a

x y b

Cu III: 2

cos

0

sin .sin 2 x I e x xdx

Cu IV nh chp S.ABCD c nh vung c (ABCD)v SA = a. G , SC. Tnh th kho

Cu V2

cos 2 , .2

x xe x x x R

II. PH nh chu

Cu VI.a 1) Trong m nhA(1; 2) v c n (C) c ph nh 2 2( 2) ( 1) 25 x y theo m

i b2) Trong khng gian v Oxyz, cho m S nh

011642222 z y x z y x v m ) c p nh 2 x + 2 y z + 17 =0. Vi nh m ) song song v ) v c S) theo giao tuy

n c chu vi b .Cu VII.a n c 5 ch

7}. Hy tnh xc su n chia h

nh nng caoCu VI.b 1) Trong m ABC bi

nh d 1: 3x nh d 2: x + 2y 5= 0. Tm to2) Trong khng gian v 1; 1; 0), B(1; 1; 2), C(2; 2; 1), D(1;1;1). Vi nh m

tr cCu VII.b 0 1 2 10042009 2009 2009 2009...S C C C C


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s 18

I. PHCu I: m s 2 3

2 x

y x

1) Kh n v v C ) c m s2) Cho M trn ( C ). Ti C) t M c(C ) t A v B. G I m to M sao cho

n ngo IAB c diCu II (2

1) Gi nh: 2 21 sin sin cos sin 2cos2 2 4 2 x x x

x x

2) Gi nh: 22 12

1log (4 4 1) 2 2 ( 2) log

2 x x x x x

Cu III 21

ln3 ln

1 ln

e x I x x dx

x x

Cu IV nh chp S.ABC c AB = AC = a . BC =2a . 3SA a , 030SAB SAC

Tnh th S.ABC .

Cu V Cho a , b, c l ba s n : a + b + c = 34

. Tm gi tr

c3 3 3

1 1 1

3 3 3P

a b b c c a.

II. PH NG ( )A. T nh Chu

Cu VIa 1) Trong m Oxy, 1 : 2 5 0d x y .d2: 3x + 6y 7 = 0. L nh P ( 2; 1) sao choth d 1 v d 2 t

d 1, d 2.2) Trong khng gian v Oxyz, A ( 1; 1; 2), B( 1; 3; 2), C ( 4;3; 2), D ( 4; 1; 2) v m P trnh: 2 0 x y z . G A l hnhchi A ln m Oxy . G S) l m A , B, C, Dto bn knh c n ( C ) l giao c P ) v ( S).

Cu VIIa nh ph h2 4 y x x v 2 y x .

B. nh Nng caoCu VIb

1) Trong m Oxy, cho Hypebol ( H nh:2 2

116 9 x y . Vi g trnh chnh t E ng v

c H ) v ngo nh ch H ). 2) Trong khng gian v Oxyz, cho : 2 5 0P x y z

3( ) : 1 3

2 x

d y z A( 2; 3; 4). G n ( P

d ) v ( P d . Tm trn M sao cho khocch AM ng

Cu VIIb nh3 1 2 3

2

2 2 3.2 (1)

3 1 1 (2)

x y y x

x xy x.


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s 19

I. PH UNG CHO TCu I m s 3 23 4 y x x .

1) Kh n v v C ) c m s2) G d A(3; 4) v c h m. Tm m d c C ) t

A, M , N sao cho hai ti C ) t M v N vung gc vCu II

1) Gi nh:2

2

1 ( ) 4

( 1)( 2)

x y x y y

x x y y( x, y R )

2) Gi nh:3 3sin .sin 3 cos cos3 1

8tan tan6 3

x x x x

x x

Cu III h tch phn:1

2

0

ln( 1) I x x x dx

Cu IV nh l ABC . A BC a , hnh chivung gc c A ln m ABC ) trng v O c ABC . Mph P ) ch BC v vung gc v AA, c

b2 38

a . Tnh th ABC . A BC .

Cu V Cho a , b, c l ba s n abc = 1. Tm gi trbi 2 2 2 2 2 2

1 1 12 3 2 3 2 3

Pa b b c c a

II. PH nh chu

Cu VI.a1) Trong m nhtrung tuy 2 1 0 x y v phn gic trong CD: 1 0 x y . Vi nh

2) Trong khng gian v nh tham s2 ; 2 ; 2 2 x t y t z t . G 1) song song

v I(2;0;2) l hnh chi n (D). Vi nh cph v c kho l

Cu VII.a Tm h x2 trong khai tri

4

1

2

n

x x , bi n l s n:2 3 1

0 1 22 2 2 656022 3 1 1

nn

n n n nC C C C n n( k nC l s k c n ph

nh nng caoCu VI.b

1) Trong m Oxy, d 1: x + y + 5 = 0, d 2: x + 2 y 7= 0 v tam gic ABC c A(2; 3), tr G B thu d 1 v

C thu d 2 . Vi nh n ngo ABC .2) Trong khng gian v Oxyz, cho tam gic ABC v A(1; 2; 5), B(1; 4;3), C (5; 2; 1) v m P ): x y z 3 = 0. G M l m n m

ph P ). Tm gi tr2 2 2

MA MB MC .Cu VII.b nh 2( 1)

1

x y x y

x y

e e x

e x y( x, y R ).


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s 20

I. PHCu I. ( ) Cho hm s 3 2( ) 3 4 f x x x .

1) Kh n v v m s

2) Tm gi tr nh m s =3 2

1 12sin 3 2sin 4

2 2 x x

Cu II. ( )1) Tm m nh sau c nghi ln( ) 2ln( 1)mx x

2) Gi nh: 3 3sin .(1 cot ) cos (1 tan ) 2sin 2 x x x x x.

Cu III. ( ) Tnh gi2

0

2 1lim

3 4 2

x

x

e x

x x

Cu IV. ( i bn knh c ABCD c 2, 3, 1, 10, 5, 13 AB AC AD CD DB BC .

Cu V. ( ) Tm m nh sau c nghi 2 : x 2 2

3

3 5

x y

x y m

II. PH NG ( ) nh Chu

Cu VI.a ( )1) Trong m nh n n

ABC v 2;3), 1 ;0 , (2;0)4

B C .

2) Trong khng gian v to nh d 4; 5;3 M v c

2 3 11 0' :

2 7 0

x yd

y zv 2 1 1'':

2 3 5 x y z

d .

Cu VII.a ( ) Tm n sao cho 1 2 3 26 6 9 14n n nC C C n n k nC l sk t n ph

nh Nng caoCu VI.b ( )

1) Trong m nh elip v1 21;1 , 5;1F F v tm sai 0,6e .

2) Trong khng gian v nh hnh chi

2 0:3 2 3 0 x zd x y z

trn m : 2 5 0P x y z .

Cu VII.b ( ) V n m k sao cho 2 2n nn k n k C C lho


21/35


s 21

I. PHCu I: m s (C m)

1) Kh n v v m s trn khi m = 1.

m cc gi trcho (d) c m) tb .

Cu II:1) Gi nh: 1 115.2 1 2 1 2 x x x

2) Tm m nh: 22 0,54(log ) log 0 x x m c nghi

Cu III:3

6 21 (1 )

dx x x

.

Cu IV: nh chp S.ABC, bi ma, m n (SAB) vung gc v n cn l ng t

Cu V: m gi tr m s 2cos

sin (2cos sin ) x

x x xv

3.

II. PHA. Theo nh chu

Cu VI.a 1) Trong m v c 3), B(3;2), ABC c di

b ; tr ABC thu y 8 = 0. Tm bn knh

n n ABC.2) Trong khng gian v

nhx 1 y 2 z 3

2 1 1. Tnh kho

Vi nh m

Cu VII.a nh2

4 3 1 02

z z z z trn t

nh nng caoCu VI.b

1) Trong m nh tin (C 1): x 2 + y2 2x 2y 2 = 0, (C 2): x 2 + y2 8x 2y + 16 = 0.


(d1) : 46 2

x t y t

z t

; v (d 2) :'

3 ' 6

' 1

x t y t

z t

G hnh chi 1; 1) trn (d 2). Tm ph nh tham sc 1) v c 1).

Cu VII.b 0 1 2 20092009 2009 2009 20092 3 ... 2010S C C C C .


22/35


s 22

I. PHCho hm s 3 23 y x x m (1)

1) Kh n v v m s m = 4.2) Tm m m s A, B sao cho 0120 . AOB

1) Gi nh: sin 3 sin 2 sin4 4

x x x .

2) Gi nh: 1 3 3 1 38 2 4 2 5 x x x .

Tnh di nh ( H ) gi 21 2 y x x v y = 1.Cho hnh chp S. ABC ABC vung cn t A, AB = AC = a . M

bn qua c BC vung gc v n cn l0. Tnh th S. ABC .

Cho a , b, c l ba s

3 2 3 2 3 2 6ab bc ca a b c

a b c b c a c a b

II. PH nh chu

1) Trong khng gian v Oxyz 1 2 2:3 2 2

x y z v

m P ): x + 3 y + 2 z + 2 = 0. L nhph P M (2; 2; 4) v c ).

2) Trong m Oxy, A(1; 0), B(3;( ): x 2 y 1 = 0. Tm C thu ) sao cho di ABC b

Tm cc s nh 0 nh1 lm m

nh nng cao

1) Trong m nh ch

tm I thu ( ) : 3 0d x y 92 I

x

c tr m t nh ch2) Trong khng gian v mtrnh l 2 2 2( ) : 4 2 6 5 0, ( ) : 2 2 16 0S x y z x y z P x y z

i ng

Gi nh:2009

22008

(1 )2. 2 0

(1 )i

z z ii

trn t


23/35


s 23

I. PH

Cho hm s3

.1) Kh m s2) D nh: x 3 x = m 3 m

1) Gi nh: cos 2x + cosx + sin 3x = 0

2) Gi nh: 3 2 2 2 2 1 3 0 x x

.

Cho I =ln 2 3 2

3 20

2 11

x x

x x x

e edx

e e e. Tnh e I

Cho hnh chp t hnh thang vung tai Av D. Bi n SD vung gc v SD =a. Tnh th

Cho tam gic ABC. Tm gi tr

P =

2 2

2

1 tan 12 2

1 tan2

A Btan

C +

2 2

2

1 tan 12 2

1 tan2

B C tan

A+

2 2

2

1 tan 12 2

1 tan2

C Atan

B

II. PH NG: ) A. nh chu :

1) Trong m n (C): x 2 + y 2 4y 5 = 0. Hy

vi nh n (C n (C) qua 4 2;5 5


c 12

:1 3 3 x y z v 2 : 4

1 2

x t

y t

z t

.

Cho t R/ x 4 13x 2 + 36 0}. Tm gi tr gitr m s 3 3x trn D.

nh nng cao:

1) Trong m n (C) v2 2( ) : 4 2 0; : 2 12 0C x y x y x y . Tm n sao cho t

0.2) Trong khng gian v nh

1

7 3 9:

1 2 1 x y z v 2 :

3 7

1 2

1 3

x t

y t

z t

Gi nh z 3 + (1 2 i)z2 + (1 i)z 2 i = 0., bitrnh c m


24/35


s 24

I. PHCu I: ) Cho hm s 3 2(1 2 ) (2 ) 2 y x m x m x m (1) ( m l tham s

1) Kh n v v m s2) Tm cc gi tr m s

Cu II

1) Gi nh: 1cos3 cos 2 cos2

x x x

2) Gi nh: 3log 3 2log 2 3log 3 log 2

x x

x x

Cu III: Tnh tch phn:6

2 2 1 4 1

dx I x x

Cho hnh chp l i SA = a, AB = b. Tnh thc nh chp

Cho x, y l cc s n 2 2 3. x xy y

Ch 2 2(4 3 3) 3 4 3 3. x xy y II. PH

A. nh chuCu VI.a:

1) Trong m nh ccth 4x + 3y 4 = 0; x y 1 = 0. Phn gic trongc y 6 = 0. Tm t2) Trong khng gian v z + 4 = 0 v hai

K sao cho KI vung gc v mph

Cu VII.a: Ch 2010 2008 20063(1 ) 4 (1 ) 4(1 )i i i i nh nng cao

Cu VI.b:1) Trong m 5y n(C): 2 2 2 4 8 0 x y x y t n (C) v

m t n(C) sao cho tam gic ABC vung2) Trong khng gian v

1

1

( ) : 1

2

x t

y t

z

, 23 1

:1 2 1

x y z

n 1 n 2 i nhCu VII.b: Cho t u s n c 5 ch

khc nhau ch


25/35


s 25

I. PHCho hm s 3 3 (1)

1) Kh n v v m s hi m = 1.

2) Tm k nh sau c nghi

3

2 32 2

1 3 0

1 1log log ( 1) 1

2 3

x x k

x x

Cu II:1) Tm t nh: sinx cos2x = 0.2) Gi nh: 31 82

2

log 1 log (3 ) log ( 1) 0 x x x .

Cu III: i Tnh tch phn:1

2 lne

I x xdx x

.

Cho hnh chp S.ABCD c nh thoi c 060 BAD , SAvung gc mqua AC v song v BD, c nh chp l , D . Tnhth C D .

Cu V: Cho a, b, c l ba c

( ) ( ) ( )ab bc ca a b c

c c a a a b b b c c a a b b c

II. PH nh chu

1) Trong m nh hai c 5x 2y + 6 = 0 v 4x + 7y 21 = 0. Vi nh cbi ng v2) Trong khng gian v nh mph A l tr IJK.

Tnh t 2 3 2525 25 251.2. 2.3. ... 24.25.S C C C . nh nng cao

Cu

1) Trong m n (C): x2

+ y2

6x + 5 = 0. TmM thu gc gituy 0.2) Trong khng gian vC(0;2;0); D(3;0;0). Vi nhv c

Tm s n 5 z v phph


26/35


s 26

I. PH

Cu I: Cho hm s21

x y x .

1) Kh n v v m s2) Ch x + m lun cth m gi tr

1) Gi nh: 41

log 2 log 02 x

x

2) Gi nh: tan tan .sin 3 sin sin 26 3

x x x x x

Cu III: Tnh tch phn2

30

sinsin 3 cos

xdx x x

Cu IV: Tnh th nh chp S.ABC bi 060 ASB ,0 090 , 120 BSC CSA .

Cu V: V n + c = 1. Tm gi tr

nh3 3 3

2 2 2(1 ) (1 ) (1 )a b c

Pa b c

II. PH nh chu :

Cu VI.a:1) Trong m 1): x + y + 1 = 0,(d2): 2x y 1 = 0 . L nh 1) c 1) v (d 2)

B sao cho 2 0 MA MB 2) Trong khng gian v 2z + 1 = 0 v

1), B(4;2;0). L nh hnh chigc c n (P).

Cu VII.a: K hi 1 v x 2 l hai nghi nh 2x 2 2x + 1 = 0.Tnh gi tr 2

1

1 x

v 22

1 x

.

B. Theo c nh nng cao:Cu VI.b:

1) Trong m nh2 2

19 4

x y . Gi m F l m

(H), k (d). Ch lun n n m n c nh n

2) Trong khng gian vTm to

Cu VII.b: Ch tho n ta lun c:.


27/35


s 27

I. PHCu I m s 4 2(2 1) 2 y x m x m (m l tham s

1) Kh n v v m s

2) Tm t m s Cu II

1) Gi nh: 2 21 8 21 1

2cos os 3 sin 2( ) 3cos s in x3 3 2 3

x c x x x .

2) Gi trnh:

1 2

2

(1 4 ).5 1 3 (1)

13 1 2 (2)

x y x y x y

x y y y x

.

Cu III nh ph : 20, , 11

x xe y y x

x

.

Cu IV nh chp S.ABCD c nh thang AB=BC=a,090 BAD , 2SA a , tam gic SCD vung t hnh chi

A trn SB. Tnh th v kho mp(SCD).

Cu V Cho x, y, z l cc s n 1 1 1 2009 x y z

. Tm gi tr

c1 1 1

2 2 2 x y z x y z x y z

II. PH

nh chu Cu VI.a1) Trong khng gian v (4; 0; 0) , (0; 0; 4) A B v m

ph 2 2 4 0 x y z . Tm n m2) Trong m d: x 5y n

(C): 2 2 2 4 8 0 x y x y t i n (C) v

m t n (C) saocho tam gic ABC vungCu VII.a m ph (1 ) n z i N v th n:

4 5log 3 log 6 4n n

nh nng caoCu VI.b (2


1 2

24 1 5

: v : d : 3 3 .3 1 2

x t x y z

d y t t

z t

Vi nh m 1 v d 2.2) Trong m nh bnh hnh ABCD c diBi Tm

t D.Cu VII.b 1 3. z i . Hy vi zn n N v th n: 23 3log ( 2 6) log 52 22 6 4 ( 2 6)n nn n n n .


28/35


s 28

I. PHCu I m s 4 25 4, y x x

1) Kh n v v2) Tm m nh 4 2 2| 5 4 | log x x mc 6 nghi

Cu II

1) Gi nh: 1 1sin 2 sin 2cot 22sin sin 2

x x x x x

2) Tm m nh: 2 2 2 1 (2 ) 0m x x x x c nghi 0; 1 3

Cu III4

0

2 1

1 2 1

x I dx

x

Cu IV 1B1C1 c AB = a, AC = 2a, AA 1 2 5a v120 o BAC . G 1. Tnh kho

ph 1BM).Cu V Cho x, y, z l cc s 3 2 4 3 5 x y z xy yz zx II. PH

A. Theo c nh chuCu VI.a.

1) Trong khng gian v 1; 3; 2), B(3; 7; 18) vm y + z + 1 = 0. Tm t (P) sao cho MA + MB nhnh2) Trong m ng trnhM(3;1) v c B v C sao cho tam gic ABC cn tA(2;2).

Cu VII. nh: 2 23 3log 1 log 2 x x x x x nh nng cao

Cu VI.b. 1) Trong khng gian v

th nh tham s1 2

1

2

x t

y t

z t

. M .

2) Trong m nhM(4;1) v c B sao cho gi tr OA OB nhnh

Cu VII.b nh: 24 2(log 8 log ) log 2 0 x x x


29/35


s 29

I. PHCho hm s 4 2 22 (1).

1) Kh n v v m s 2.2) Tm m m s nh m

0120 .

1) Gi nh: 23 1 1 2 3 4 x x x x

2) Gi nh:2sin

4 (1 sin 2 ) 1 tancos

x x x

x

m) Tnh di nh ph , 0, 0, .1 sin

x y y x x

x

Cho hnh h B C D nh vung, AB = AA =2a. Hnh chi ln m ng v trung

nh h cosin c A C

Tm gi tr 3 25 9 4II. PH

nh chu

1) Trong m nh bnh hnh ABCD c diBi BD n

2) Trong khng gian v Oxyz, cho A(2; 0; 0), B(0; 2; 0), C(0; 0; 2). Tnh bnknh m

Ch 0 10 1 9 9 1 10 0 1010 20 10 20 10 20 10 20 30. . ... . .C C C C C C C C C . nh nng cao

1) Trong m n (C): 2 2 2 4 5 0 vA(0; 1) (C). Tm to n (C) sao cho2) Trong khng gian v 2 2 1 0 x y z v cc

1 2

1 3 5 5: ; :2 3 2 6 4 5

x y z x y zd d . Tm c 1 2d , d M N

sao cho MN // (P) v cch (P) m

Tm cc s n:1 1 1

1 1

10 2 1

y y y y x x x x A yA A C .


30/35


s 30

I. PH

Cu I. Cho hm s3 2 33 1

2 2 y x mx m 1) Kh n v v m s

m s au qua

1) Gi nh: 2 2 3 3tan tan .sin cos 1 0 x x x x

2) Gi nh: 2 1 1 15.3 7.3 1 6.3 9 0 x x x x

Tnh tch phn: I =4 3

41

1( 1)

dx x x

Cho hnh chp S.ABC c m nSA vung gc v 0, tnh thS.ABC theo a.

Cho ba s3 3 3

2 2 2 2 2 2 1a b c

a ab b b bc c c ca a

Tm gi tr a + b + c II. PH

nh chu

1) Trong khng gian v nh mgc v 0 x y z v 1 ) m 2 . 2) Trong m nhgic trong gc A l (d 1 nh (d2): 2x y +1 = 0, c 1). Tm ph nh c

C 6 h 3 h ng d o lnhiu cch x

nh nng cao

1) Trong khng gian v2 4

3 2

3

x t

y t

z t

v m

ph 2 5 0 x y z . Vi nh ) n

song v cch (d) m 14 .

2) Trong m 2 m to

(P) sao cho 4 .

Tm m nh sau c nghi 25 1 5 6 x x x x m


31/35


s 31

I. PHCu I: Cho hm s 3 + 3x 2 m); (m l tham s

1) Kh n v v m s2) Xc m) ccc ti m) t E vung gc v

Cu II 1) Gi nh: 2cos3x + 3 sinx + cosx = 0

2) Gi nh:2 2

2 2

91 2 (1)

91 2 (2)

x y y

y x x

Cu III2

ln .ln

e

e

dx x x ex

Cu IV: nh chp S.ABC, bi ma, m n (SAB) vung gc v n cn l ng t

Cu V: , ,a b c l nh n: 2 2 2 3a b c . Ch

2 2 21 1 1 4 4 4

7 7 7a b b c c a a b c

II.PHA. nh chu

Cu VI.a 1) Trong m ph 2 24 9 36 x y Vi nh c2) Trong khng gian v m trn Ox

(d) : 1 21 2 2

x y z v m y 2z = 0.

Cu VII.a 0,1,2,3,4,5,6,7 . C th u s ng n ph1.B. nh nng cao

Cu VI.b 1) Trong m 2 25 16 80 x y 5; 1), B(1; 1). M n (E). Tm gi tr MAB.2) Trong khng gian v

nh (P): 3 12 3 5 0 x y z v (Q): 3 4 9 7 0 x y z

(d1):5 3 1

2 4 3 x y z , (d 2):

3 1 22 3 4

x y z .

Vi nh ) song song v c 1), (d 2)Cu VII.b Tm s n b nh: 3 22 9nn n A C n.


32/35


s 32

I. PH CCu I: m s 2 1

1 x

x.

1) Kh n v v m s2) G a. Tituy Q. Chc tnh di

Cu II 1) Gi nh: 2 2log ( 3 1 6) 1 log (7 10 ) x x

2) Gi nh:6 6

2 2

sin cos 1tan2

cos sin 4 x x

x x x

Cu III: phn: I =4

20

21 tan

x x ee x dx

x

Cu IV nh l m nh thoic BAD = 60 0. G . Chminh r , M, N y tnh i cgic B MDN l hnh vung.

Cu V m gi trbi 1 1 1

1 1 1P

a b c

II. PH trnh chuCu VI.a

1) Trong m trnh 2x y + 3 = 0. L nh ) qua A v t

1

10.

2) Trong khng gian v 1;3;1).L nh c c tm n n my 2z + 4 = 0.

Cu VII.a hu s n c 5 ch ph 2.

nh nng caoCu VI.b

1) Trong m ):3x 4y + 8 = 0. L nh n qua A, B v ti ).2) Trong khng gian v 1;3;1). Ch tm trABC.

Cu VII.b Gi nh: log log2 2 3

y x

x y

xy y.


33/35


s 33

I. PHCu I m s 4 3 22 3 1 (1) y x mx x mx .

1) Kh n v v m sm s

Cu II

1) Gi nh: cos3xcos 3x sin3xsin 3x = 2 3 28

2) Gi nh: 2 22 1 2 ( 1) 2 3 0 x x x x x x

Cu III2

0

1 sin 2 I x xdx.

Cu IV: (1 .ABC l hnh chp tam gicAB = a, c n AA = b. G l gc gi (A BC). Tnh

tan v th .BB C C.Cu V:

2 2 2

2 2 2

a b c a b cb c a b c a

.

II. PH nh chu

Cu VI.a: 1) Trong m nh ch

trung: x + y 5 = 0. Vi nh

th2) Trong khng gian v 2y z 4 = 0 v mc 2 + y 2 + z 2 2x 4y 6z 11 = 0. Chc n. Xc tnh bn knh c n

Cu VII.a: nh: 2 21 29 1 10.3 x x x x . nh nng cao

Cu VI.b: 1) Trong m n (C): x 2 + y 2 + 4x + 4y + 6 = 0 v

: x + my 2m + 3 = 0 v tham s tm c n(C). Tm m c B sao cho di IAB l2) Trong khng gian v nh mD(1; 1; 1) v c tr

cCu VII.b: nh: 14 2 2(2 1)sin(2 1) 2 0 x x x x y .


34/35


s 34

I. PHCu I: ): Cho hm s 4 22 1.

1) Kh n v v m s

2) Bi trnh: 4 2 22 1 log 0 x x m (m>0)

Cu II: ) 1) Gi nh: 2 23 2 2 3 1 1 x x x x x

2) Gi nh : 3 3 2cos cos 3 sin sin 3 4 x x x x

Cu III: ): Tnh tch phn: I=2

30

7sin 5cos(sin cos )

x xdx

x x

Cu IV: ): Cho hnh chp t i cbn t o. MSAC c

Cu V: a, b, c, d tho n: 2 2 1 ; c d = 3.

Ch 9 6 24

F ac bd cd

II.PH nh Chu

Cu VI.a:1) Trong m 7), B(9; 5), C(5;9), M(2; 7). Vi nh ti nngo ABC.2) Trong khng gian v

1 :x y z

d 1 1 2

v 2

1 2

:

1

x t

d y t

z t

Xt v 1 v d 2. Vi nh 2 v vunggc v 1

Cu VII.a: n bi tr 7 vin bi vng. Nguch n bi t u cch ch bi l

u? nh Nng cao :

Cu VI.b:1) Trong mtrung tuy nh l: x 2y + 1 = 0 v y 1 = 0. Hy vi gtrnh cc c ABC.2) Trong khng gian v A(0; 0;3), B(2; 0;1) v mph P nh: 3 8 7 1 0 x y z . Vi nh chnh td n n m P) v d vung gc v AB t AB v(P).

Cu VII.b: m h 3 trong khai tri 2 2n

x x

bi n:

1 3 2 1 232 2 2... 2nn n nC C C


35/35


s 35

I. PH

Cu I Cho hm s x 2

2x 3

(1).

1) Kh n v v m s 2) Vi nh ti m s nh,tr sao cho OAB cn t

Cu II1) Gi nh: cot 3 tan 2cot 2 3 x x x .

2) Gi nh: 2 22( 1) 3 1 2 2 5 2 8 5 x x x x x x .

Cu III Tnh tch phn :4

0

cos sin

3 sin 2

x x I dx

x.

Cu IV Cho hnh l B C D c trungD = 2PD. Chvung gc v AM) v tnh th AMP.

Cu V 3 c vi b . Tm gi tr

bi3 3 3( ) ( ) ( )

3 3 3a b c b c a c a b

Pc a b

.

II. PH nh chu

Cu VI.a.1) Trong m n (C): (x 1) 2 + (y + 1) 2 = 25 v

nhcho MA = 3MB.2) Trong khng gian v 2y + 2z 1 = 0 v hai

1 :x 1 y z 9

1 1 6; 2 :

x 1 y 3 z 12 1 2

. X

M thu 1 sao cho kho 2 v khot

Cu VII.a G 1 v z 2 l 2 nghi nh: 2 2 10 0 .Tnh gi tr bi 2 21 2 A z z .

nh nng cao

Cu VI.b1) Trong m 1), C(11;2). Vi nh chia ABC thnh hai ph stch b

2) Trong khng gian v 1 2:1 2 1 x y z

d v m

ph nh4), song song v c

Cu VII.b nh: 32 7log 1 log x x .

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