BỘ ĐỀ THI THỬ ĐẠI HỌC 2009 CHỌN LỌC (toanhoccapba.wordpress.com)

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THI THI HC 2009 CHN LC

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s 1Cu1: (2,5 im)

Cho hm s: y = -x3 + 3mx2 + 3(1 - m2)x + m3 - m21) Kho st s bin thin v v th ca hm s trn khi m = 1.2) Tm k phng trnh: -x3 + 3x2 + k3 - 3k2 = 0 c 3 nghim phn bit.

3) Vit phng trnh ng thng i qua 2 im cc tr ca th hm s trn.Cu2: (1,75 im)

Cho phng trnh: 01212323 =++ mxlogxlog (2)

1) Gii phng trnh (2) khi m = 2.

2) Tm m phng trnh (2) c t nht 1 nghim thuc on

331; .

Cu3: (2 im)

1) Tm nghim (0; 2) ca pt : 32221

335 +=

+++ xcos

xsinxsinxcos

xsin

2) Tnh din tch hnh phng gii hn bi cc ng: y = 342 + xx , y = x + 3

Cu4: (2 im)1) Cho hnh chp tam gic u S.ABC nh S c di cnh y bng a. Gi M

v N ln lt l trung im ca cc cnh SB v SC. Tnh theo a din tch AMN bitrng mt phng (AMN) vung gc mt phng (SBC).

2) Trong khng gian Oxyz cho 2 ng thng: 1:

=++=+

0422

042

zyx

zyx

v 2:

+=+=+=

tz

ty

tx

21

2

1

a) Vit phng trnh mt phng (P) cha ng thng 1 v song song vi ngthng 2.

b) Cho im M(2; 1; 4). Tm to im H thuc ng thng 2 sao cho onthng MH c di nh nht.Cu5: (1,75 im)

1) Trong mt phng vi h to cc vung gc Oxy xt ABC vung ti

A, phng trnh ng thng BC l: 033 = yx , cc nh A v B thuc trchonh v bn knh ng trn ni tip bng 2. Tm to trng tm G ca ABC

2 Khai trin nh thc:


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nxnn

nxxnn

xnx

n

nx

n

nxx

CC...CC

+

++

+

=

+

3

1

32

1

13

1

2

1

12

1

032

1

22222222

Bit rng trong khai trin 13 5 nn CC = v s hng th t bng 20n, tm n v x

s 2Cu1: (2 im)

Cu Cho hm s: y = mx4 + (m2 - 9)x2 + 10 (1)

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

2) Tm m hm s (1) c ba im cc tr.

Cu2: (3 im)

1) Gii phng trnh: sin23x - cos24x = sin25x - cos26x

2) Gii bt phng trnh: logx(log3(9x - 72)) 1

3) Gii h phng trnh:

++=+

=

2

3

yxyx

yxyx

Cu3: (1,25 im)

Tnh din tch hnh phng gii hn bi cc ng: y =x

yvx 2

2444

2

=

Cu4: (2,5 im)1) Trong mt phng vi h to cc vung gc Oxy cho hnh ch nht

ABCD c tm I

0

2

1; , phng trnh ng thng AB l x - 2y + 2 = 0 v AB = 2AD.

Tm to cc nh A, B, C, D bit rng nh A c honh m

2) Cho hnh lp phng ABCD.A1B1C1D1 c cnh bng a

a) Tnh theo a khong cch gia hai ng thng A1B v B1D.

b) Gi M, N, P ln lt l cc trung im ca cc cnh BB1, CD1, A1D1. Tnh gcgia hai ng thng MP v C1N.

Cu5: (1,25 im)

Cho a gic u A1A2...A2n (n 2, n Z) ni tip ng trn (O). Bit rng s

tam gic c cc nh l 3 im trong 2n im A1, A2, ... ,A2n nhiu gp 20 ln s hnh

ch nht c cc nh l 4 im trong 2n im A1, A2, ... ,A2n . Tm n.


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s 3Cu1: (3 im)

Cho hm s: y =( )

1

122

xmxm

(1) (m l tham s)

1) Kho st s bin thin v v th (C) ca hm s (1) ng vi m = -1.

2) Tnh din tch hnh phng gii hn bi ng cong (C) v hai trc to .

3) Tm m th ca hm s (1) tip xc vi ng thng y = x.

Cu2: (2 im)

1) Gii bt phng trnh: (x2 - 3x) 0232 2 xx .

2) Gii h phng trnh:

=+

+

=+

y

yy

x

xx

x

22

24

452

1

23

Cu3: (1 im)

Tm x [0;14] nghim ng phng trnh: cos3x - 4cos2x + 3cosx - 4 = 0 .Cu4: (2 im)

1) Cho hnh t din ABCD c cnh AD vung gc vi mt phng (ABC); AC =

AD = 4 cm ; AB = 3 cm; BC = 5 cm. Tnh khong cch t im A ti mt phng

(BCD).

2) Trong khng gian vi h to cc vung gc Oxyz, cho mt phng

(P): 2x - y + 2 = 0 v ng thng dm:

( ) ( )

( )

=++++

=+++02412

01112

mzmmx

mymxm

Xc nh m ng thng dm song song vi mt phng (P) .

Cu5: (2 im)

1) Tm s nguyn dng n sao cho: 243242 210 =++++ nnn

nnn C...CCC .


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2) Trong mt phng vi h to cc vung gc Oxy cho Elp (E) c

phng trnh: 1916

22

=+ yx . Xt im M chuyn ng trn tia Ox v im N chuyn

ng trn tia Oy sao cho ng thng MN lun tip xc vi (E). Xc nh to ca

M, N on MN c di nh nht. Tnh gi tr nh nht .

s 4

Cu1: (2 im)

Cho hm s: y =1

32

+

xx

1) Kho st s bin thin v v th hm s.

2) Tm trn ng thng y = 4 cc im m t k c ng 2 tip tuyn

n th hm s.Cu2: (2 im)

1) Gii h phng trnh:

=++

=++

0

123

yxyx

yxyx

2) Gii bt phng trnh: ( ) 012

1 2 >++ xxlnxln

Cu3: (2 im)

1) Gii phng trnh: cosx+ cos2x + cos3x + cos4x + cos5x = -2

1

2) Chng minh rng ABC tho mn iu kin

224

22

2

7 Bcos

Acos

CsinCcosBcosAcos ++=+ th ABC u

Cu4: (2 im)

1) Trn mt phng to cho A(1, 0); B(0, 2); O(0, 0) v ng trn (C) c

phng trnh: (x - 1)2 +2

21

y = 1. Vit phng trnh ng thng i qua cc giao

im ca ng thng (C) v ng trn ngoi tip OAB.


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2) Cho hnh chp S.ABC c y ABC l tam gic vung cn vi AB = AC = a,

SA = a, SA vung gc vi y. M l mt im trn cnh SB, N trn cnh SC sao cho

MN song song vi BC v AN vung gc vi CM. Tm t sMBMS

.

Cu5: (2 im)

1) Tnh din tch phn mt phng gii hn bi cc ng cong: y = x3 - 2 v

(y + 2)2 = x.

2) Vi cc ch s 1, 2, 3, 4, 5, 6 c th lp c bao nhiu s c 3 ch s khc

nhau, bit rng cc s ny chia ht cho 3.

s 5

Cu1: (2 im)

Cho hm s: y = x + 1 +1

1x

.

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

2) T mt im trn ng thng x = 1 vit phng trnh tip tuyn n th (C).

Cu2: (2 im)

1) Gii phng trnh: 1635223132 2 +++=+++ xxxxx

2) Tm cc gi tr x, y nguyn tho mn: ( ) yyxxlogy

3732

282

2

2

+++

+

Cu3: (2 im)

1) Gii phng trnh: (cos2x - 1)(sin2x + cosx + sinx) = sin22x

2) ABC c AD l phn gic trong ca gc A (D BC) v sinBsinC 2

2 Asin .

Hy chng minh AD2 BD.CD .

Cu4: (2 im)

1) Trn mt phng to vi h to cc vung gc Oxy, cho elip cphng trnh: 4x2 + 3y2 - 12 = 0. Tm im trn elip sao cho tip tuyn ca elip ti

im cng vi cc trc to to thnh tam gic c din tch nh nht.

2) Trong khng gian vi h trc to cc vung gc Oxyz, cho hai mt

phng (P): x - y + z + 5 = 0 v (Q): 2x + y + 2z + 1 = 0. Vit phng trnh mt cu c

tm thuc mt phng (P) v tip xc vi mt phng (Q) ti M(1; - 1; -1).


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Cu5: (2 im)

1) Tnh din tch hnh phng gii hn bi cc ng: y = 2 -4

2xv x + 2y = 0

2) a thc P(x) = (1 + x + x2)10 c vit li di dng: P(x) = a0 + a1x + ... +

a20x20. Tm h s a4 ca x4.

s 6

Cu1: (2 im)

Cho hm s: y =1

2

++

xmxmx (1) (m l tham s)

1) Kho st s bin thin v v th ca hm s (1) khi m = -1.

2) Tm m th hm s (1) ct trc honh ti hai im phn bit v hai im

c honh dng.

Cu2: (2 im)

1) Gii phng trnh: cotgx - 1 =tgx

xcos+1

2+ sin2x -

2

1sin2x

2) Gii h phng trnh:

+=

=

12

11

3xy

yy

xx

Cu3: (3 im)

1) Cho hnh lp phng ABCD.A'B'C'D'. Tnh s o ca gc phng nh din

[B, A'C, D].

2) Trong khng gian vi h to cc Oxyz cho hnh hp ch nht

ABCD.A'B'C'D' c A trng vi gc ca h to , B(a; 0; 0), D(0; a; 0), A'(0; 0; b)(a > 0, b > 0). Gi M l trung im cnh CC'.

a) Tnh th tch khi t din BDA'M theo a v b.

b) Xc nh t sba

hai mt phng (A'BD) v (MBD) vung gc vi nhau.

Cu4: (2 im)


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1) Tm h s ca s hng cha x8 trong khai trin nh thc Niutn ca:n

xx

+ 53

1, bit rng: ( )373

14 += +

++ nCC

nn

nn (n N*, x > 0)

2) Tnh tch phn: I = +

32

52

4xx

dx

Cu5: (1 im)

Cho x, y, z l ba s dng v x + y + z 1. Chng minh rng:

821112

2

2

2

2

2 +++++z

zy

yx

x

s 7Cu1: (2 im)

Cho hm s: y = x3 - 3x2 + m (1)1) Tm m th hm s (1) c hai im phn bit i xng vi nhau qua gc

to .2) Kho st s bin thin v v th ca hm s (1) khi m = 2 .

Cu2: (2 im)

1) Gii phng trnh: cotgx - tgx + 4sin2x =xsin2

2

2) Gii h phng trnh:

+=

+

=

2

2

2

2

23

2

3

y

xx

x

y

y

Cu3: (3 im)

1) Trong mt phng vi h ta cc vung gc Oxy cho ABC c: AB =

AC, = 900. Bit M(1; -1) l trung im cnh BC v G

0

3

2; l trng tm ABC.

Tm to cc nh A, B, C .2) Cho hnh lng tr ng ABCD.A'B'C'D' c y ABCD l mt hnh thoi cnh a,

gc = 600 . gi M l trung im cnh AA' v N l trung im cnh CC'. Chngminh rng bn im B', M, D, N cng thuc mt mt phng. Hy tnh di cnhAA' theo a t gic B'MDN l hnh vung.


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3) Trong khng gian vi h to cc Oxyz cho hai im A(2; 0; 0) B(0; 0; 8)

v im C sao cho ( )060 ;;AC = . Tnh khong cch t trung im I ca BC n ngthng OA.Cu4: (2 im)

1) Tm gi tr ln nht v nh nht ca hm s: y = x +2

4 x

2) Tnh tch phn: I =

+4

0

2

21

21dx

xsinxsin

Cu5: (1 im)Cho n l s nguyn dng. Tnh tng:

nn

n

nnn Cn

...CCC1

12

3

12

2

121

23

12

0

+++++

+

( knC l s t hp chp k ca n phn t) s 8

Cu1: (2 im)

1) Kho st s bin thin v v th ca hm s: y =2

422

+

xxx

(1)

2) Tm m ng thng dm: y = mx + 2 - 2m ct th ca hm s (1) ti hai

im phn bit.

Cu2: (2 im)1) Gii phng trnh: 0

242

222 =

xcosxtgxsin

2) Gii phng trnh: 32222

2 = + xxxx Cu3: (3 im)

1) Trong mt phng vi h ta trc cc vung gc Oxy cho ng trn:

(C): (x - 1)2 + (y - 2)2 = 4 v ng thng d: x - y - 1 = 0

Vit phng trnh ng trn (C') i xng vi ng trn (C) qua ng thng d.

Tm ta cc giao im ca (C) v (C').2) Trong khng gian vi h to cc vung gc Oxyz cho ng thng:

dk:

=++=++01

023

zykx

zkyx

Tm k ng thng dk vung gc vi mt phng (P): x - y - 2z + 5 = 0.


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3) Cho hai mt phng (P) v (Q) vung gc vi nhau, c giao tuyn l ng

thng . Trn ly hai im A, B vi AB = a. Trong mt phng (P) ly im C, trong

mt phng (Q) ly im D sao cho AC, BD cng vung gc vi v AC = BD = AB.Tnh bn knh mt cu ngoi tip t din ABCD v tnh khong cch t A n mt

phng (BCD) theo a.

Cu4: (2 im)

1) Tm gi tr ln nht v gi tr nh nht ca hm s: y =1

1

2 +

+

x

x

trn on [-1; 2]

2) Tnh tch phn: I = 2

0

2 dxxx

Cu5: (1 im)

Vi n l s nguyn dng, gi a3n - 3 l h s ca x3n - 3 trong khai trin thnh athc ca (x2 + 1)n(x + 2)n. Tm n a3n - 3 = 26n.

s 9

Cu1: (2 im)

Cho hm s: y =( )12

332

+

xxx

(1)

1) Kho st s bin thin v v th ca hm s (1).

2) Tm m ng thng y = m ct th hm s (1) ti hai im A, B sao cho

AB = 1.

Cu2: (2 im)

1) Gii bt phng trnh:( )

3

73

3

1622

>+

x

xx

x

x

2) Gii h phng trnh:

( )

=+

=

25

11

22

4

4

1

yxy

logxylog

Cu3: (3 im)

1) Trong mt phng vi h ta cac Oxy cho im A(0; 2) v B ( )13 ; .

Tm to trc tm v to tm ng trn ngoi tip OAB.


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2) Trong khng gian vi h to cc Oxyz cho hnh chp S.ABCD c y

ABCD l hnh thoi, AC ct BD ti gc to O. Bit A(2; 0; 0) B(0; 1; 0)

S(0; 0; 2 2 ). Gi M l trung im ca cnh SC.

a) Tnh gc v khong cch gia hai ng thng SA v BM.

b) Gi s mt phng (ABM) ct SD ti N. Tnh th tch hnh chp S.ABMN.Cu4: (2 im)


2

1 11dx

x

x

2) Tm h s ca x8 trong khai trin thnh a thc ca: ( )[ ]82 11 xx + Cu5: (1 im)

Cho ABC khng t tho mn iu kin: cos2A + 2 2 cosB + 2 2 cosC = 3

Tnh cc gc ca ABC.

s 10Cu1: (2 im)

Cho hm s: y = xxx 323

1 23 + (1) c th (C)


2) Vit phng trnh tip tuyn ca (C) ti im un v chng minh rng ltip tuyn ca (C) c h s gc nh nht.

Cu2: (2 im)

1) Gii phng trnh: 5sinx - 2 = 3(1 - sinx)tg2x

2) Tm gi tr ln nht v gi tr nh nht ca hm s: y =x

xln2trn on

[ ]31 e; .Cu3: (3 im)

1) Trong mt phng vi h ta cc Oxy cho im A(1; 1), B(4; -3). Tmim C thuc ng thng y = x - 2y - 1 = 0 sao cho khong cch t C n ngthng AB bng 6.

2) Cho hnh chp t gic u S.ABCD c cnh y bng a, gc gia cnh bn

v mt y bng (00 < < 900). Tnh tang ca gc gia hai mt phng (SAB) v(ABCD) theo a v .


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3) Trong khng gian vi h to cc Oxyz cho im A(-4; -2; 4) v ng

thng d:

+==

+=

tz

ty

tx

41

1

23

(t R). Vit phng trnh ng thng i qua im A, ct v

vung gc vi ng thng d.Cu4: (2 im)

1) Tnh tch phn I = +e

xdxlnx

xln

1

31

2) Trong mt mn hc, thy gio c 30 Cu hi khc nhau gm 5 Cu hi kh,

10 Cu hi trung bnh, 15 Cu hi d. T 30 Cu hi c th lp c bao nhiu kim tra, mi gm 5 Cu hi khc nhau, sao cho trong mi nht thit phi c 3 loi Cu hi (kh, d, trung bnh) v s Cu hi d khng t hn 2?

Cu5: (1 im)Xc nh m phng trnh sau c nghim:

224221112211 xxxxxm ++=

++

s 11

Cu1: (2 im)

Cho hm s y = x3 - 3mx2 + 9x + 1 (1) (m l tham s)

1) Kho st s bin thin v v th ca hm s (1) khi m = 2.2) Tm m im un ca th hm s (1) thuc ng thng y = x + 1.

Cu2: (2 im)

1) Gii phng trnh: ( )( ) xsinxsinxcosxsinxcos =+ 2212

2) Tm m h phng trnh sau:

=+

=+

myyxx

yx

31

1c nghim.

Cu3: (3 im)

1) Trong mt phng vi h ta cc Oxy cho ABC c cc nh A(-1; 0);

B(4; 0); C(0; m) vi m 0. Tm to trng tm G ca ABC theo m. Xc nh m

GAB vung ti G.

2) Trong khng gian vi h to cc Oxyz cho hnh lng tr ng

ABC.A1B1C1. Bit A(a; 0; 0); B(-a; 0; 0); C(0; 1; 0); B1(-a; 0; b) a > 0, b > 0.


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a) Tnh khong cch gia hai ng thng B1C v AC1 theo a, b.

b) Cho a, b thay i nhng lun tho mn a + b = 4. Tm a, b khong cch

gia 2 ng thng B1C v AC1 ln nht.

3) Trong khng gian vi h to cc Oxyz cho 3 im A(2; 0; 1) B(1; 0; 0)

C(1; 1; 1) v mt phng (P): x + y + x - 2 = 0. Vit phng trnh mt cu i qua 3im A, B, C v c tm thuc mt phng (P).

Cu4: (2 im)

1) Tnh tch phn I = ( ) 3

2

2 dxxxln

2) Tm cc s hng khng cha x trong khai trin nh thc Newtn ca

7

4

3 1

+ xx vi x > 0

Cu5: (1 im)

Chng minh rng phng trnh sau c ng 1 nghim: x5 - x2 - 2x - 1 = 0

s 12Cu1: (2 im)

Gi (Cm) l th ca hm s: y = mx +1

x(*) (m l tham s)

1. Kho st s bin thin v v th ca hm s (*) khi m = 14

2. Tm m hm s (*) c cc tr v khong cch t im cc tiu ca (Cm)n tim cn xin ca (Cm) bng

1

2

Cu2: (2 im)

1. Gii bt phng trnh: 5 1 1 2 4 x x x > 2. Gii phng trnh: cos23xcos2x - cos2x = 0

Cu3: (3 im)

1. Trong mt phng vi h to Oxy cho hai ng thngd1: x - y = 0 v d2: 2x + y - 1 = 0

Tm to cc nh ca hnh vung ABCD bit rng nh A thuc d1, nh Cthuc d2 v cc nh B, D thuc trc honh.


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2. Trong khng gian vi h to Oxyz cho ng thng d:1 3 3

1 2 1

x y z + = =

v mt phng (P): 2x + y - 2z + 9 = 0.

a. Tm to im I thuc d sao cho khong cch t I n mt phng(P) bng 2

b. Tm to giao im A ca ng thng d v mt phng (P). Vitphng trnh tham s ca ng thng nm trong mt phng (P),bit i qua A v vung gc vi d.

Cu4: (2 im)

1. Tnh tch phn I = 20

sin 2 sin

1 3cos

x xdx

x

++

2. Tm s nguyn dng n sao cho:( )1 2 2 3 3 4 2 12 1 2 1 2 1 2 1 2 12.2 3.2 4.2 ... 2 1 2 2005n nn n n n nC C C C n C 2 ++ + + + + + + + + =

Cu5: (1 im)

Cho x, y, z l cc s dng tho mn:1 1 1

4 x y z

+ + = . Chng minh rng:

1 1 11

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

+ + + + + +

s 13

Cu1: (2 im)

Gi (Cm) l th hm s y =( )2 1 1

1

x m x m

x

+ + + ++

(*) m l tham s

1. Kho st s bin thin v v th ca hm s (*) khi m = 1.2. Chng minh rng vi m bt k, th (Cm) lun lun c im cc i, cc

tiu v khong cch gia hai im bng 20

Cu2: (2 im)

1. Gii h phng trnh: ( )2 39 31 2 1

3log 9 log 3

x y

x y

+ = =

2. Gii phng trnh: 1 + sinx + cosx + sin2x + cos2x = 0Cu3: (3 im)


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1. Trong mt phng vi h to Oxy cho A(2; 0) v B(6; 4). Vit phngtrnh ng trn (C) tip xc vi trc honh ti hai im v khong cch t

tm ca (C) n im B bng 5.

2. Trong khng gian vi h to Oxyz cho hnh lng tr ng ABC.A1B1C1vi A(0; -3; 0) B(4; 0; 0) C(0; 3; 0) B1(4; 0; 4)

a. Tm to cc nh A1, C1. Vit phng trnh mt cu c tm l A vtip xc vi mt phng (BCC1B1).

b. Gi M l trung im ca A1B1. Vit phng trnh mt phng P) i quahai im A, M v song song vi BC1. mt phng (P) ct ng thng

A1C1 ti im N. Tnh di on MN

Cu4: (2 im)

1. Tnh tch phn: I = 20

sin 2 cos

1 cos

x xdx

x

+

2. Mt i thanh nin tnh nguyn c 15 ngi, gm 12 nam v 3 n. Hi cbao nhiu cch phn cng i thanh nin tnh nguyn v gip 3 tnh

min ni, sao cho mi tnh c 4 nam v 1 n?

Cu5: (2 im)

Chng minh rng vi mi x thuc R ta c:

12 15 203 4 5

5 4 3

x x x

x x x + + + +

Khi no ng thc xy ra?

s 14Cu1: (2 im)

Gi (Cm) l th hm s: y =3 21 1

3 2 3

mx x + (*) (m l tham s)

1. Kho st s bin thin v v th ca hm s (*) khi m = 22. Gi M l im thuc (Cm) c honh bng -1. Tm m tip tuyn ca

(Cm) ti im M song song vi ng thng 5x - y = 0Cu2: (2 im)

Gii cc phng trnh sau:

1. 2 2 2 1 1 4 x x x+ + + + = 2. 4 4 3cos sin cos sin 3 0

4 4 2 x x x x

+ + =

Cu3: (3 im)


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1. Trong mt phng vi h to Oxy cho im C(2; 0) v Elip (E):2 2

14 1

x y+ = . Tm to cc im A, B thuc (E), bit rng A, B i xng

vi nhau qua trc honh va ABC l tam gic u.2. Trong khng gian vi h to Oxyz cho hai ng thng:

d1:1 2 1

3 1 2

x y z + += =

v d2:2 0

3 12 0

x y z

x y

+ = + =

a. Chng minh rng: d1 v d2 song song vi nhau. Vit phng trnh mtphng (P) cha c hai ng thng d1 v d2

b. mt phng to Oxz ct hai ng thng d1, d2 ln lt ti cc imA, B. Tnh din tch OAB (O l gc to )

Cu4: (2 im)

1. Tnh tch phn: I = ( )2

sin

0

cos cosxe x xdx

+

2. Tnh gi tr ca biu thc M =( )

4 3

13

1 !

n nA A

n

+ ++

bit rng

2 2 2 2

1 2 3 42 2 149n n n nC C C C + + + ++ + + =

Cu5: (1 im)Cho cc s nguyn dng x, y, z tho mn xyz = 1. Chng minh rng:

3 3 3 3 3 31 1 13 3

x y y z z x

xy yz zx

+ + + + + ++ +

Khi no ng thc xy ra?

s 15Phn chung c tt c cc th sinh

Cu1: (2 im)1. Kho st s bin thin v v th ca hm s: y = 2x3 - 9x2 + 12x - 42. Tm m phng trnh sau c 6 nghim phn bit: 3 22 9 12 x x x m + =

Cu2: (2 im)

1. Gii phng trnh: ( )6 62 sin sin .cos 02 2sin

cos x x x x

x

+ =

2. Gii h phng trnh: 31 1 4

xy xy

x y

=

+ + + =


16/155



Cu3: ( 2 im) Trong khng gian vi h to Oxyz. Cho hnh lp phngABCD.ABCD vi A(0; 0; 0) B(1; 0; 0) D(0; 1; 0) A(0; 0; 1). Gi M v N ln ltl trung im ca AB v CD.

1. Tnh khong cch gia hai ng thng AC v MN.2. Vit phng trnh mt phng cha AC v to vi mt phng Oxy mt gc

bit cos = 16

Cu4: (2 im)

1. Tnh tch phn: I = 22 2

0

sin2

cos 4sin

xdx

x x

+

2. Cho hai s thc x 0, y 0 thay i v iu kin: (x + y)xy = x2 + y2 - xy.Tm GTLN ca biu thc A =

3 3

1 1

x y+

Phn T chn: Th sinh chn Cu 5.a hc Cu 5.bCu5a:Theo chng trnh khng phn ban: (2 im)

1. Trong mt phng vi h to Oxy cho cc ng thng:d1: x + y + 3 = 0 d2: x - y - 4 = 0 d3: x - 2y = 0.Tm to im M nm trn ng thng d3 sao cho khong cch t M n

ng thng d1 bng hai ln khong cch t M n ng thng d2

2. Tm h s ca s hng cha x26 trong khai trin nh thc: 74

1n

xx

+

, bit

rng: 1 2 02 1 2 1 2 1... 2 1n

n n nC C C2

+ + ++ + + = Cu5b:Theo chng trnh phn ban: (2 im)

1. Gii phng trnh: 3.8x + 4.12x - 18x - 2.27x = 02. Cho hnh lng tr c cc y l hai hnh trn tm O v O, bn knh bng

chiu cao v bng a. Trn ng trn y tm O ly im A, trn ng trn y tmO ly im B sao cho AB = 2a. Tnh th tch ca khi t din OOAB.


Cu1: (2 im)

Cho hm s: y =

21

2

x x

x

+ +

1. Kho st s bin thin v v th (C) ca hm s.2. Vit phng trnh tip tuyn ca th (C), bit tip tuyn vung gc vi

tim cn xin ca (C).Cu2: (2 im)

1. Gii phng trnh: cotx + sinx 1 tan .tan 42

xx

+ =


17/155



2. Tm m phng trnh sau c hai nghim thc phn bit:2

2 2 1 x mx x+ + = Cu3: (2 im)

Trong khng gian vi h to Oxyz cho im A(0; 1; 2) v hai ng thng :

d1:1 1

2 1 1

x y z += = d2:

1

1 2

2

x t

y t

z t

= +

= = +

1. Vit phng trnh mt phng (P) qua A, ng thi song song vi d1 v d2.2. Tm to cc im M d1, N d2 sao cho ba im A, M, N thng hng

Cu4: (2 im)

1. Tnh tch phn: I = ln5ln3

2 3x x

dx

e e+

2. Cho x, y l cc s thc thay i. Tm GTNN ca biu thc:A = ( ) ( )

2 22 2

1 1 2 x y x y y + + + + + Phn T chn: Th sinh chn Cu 5.a hc Cu 5.b

Cu5a:Theo chng trnh khng phn ban: (2 im)1. Trong mt phng vi h to Oxy cho ng trn (C): x2 + y2 -2x - 6y + 6

= 0 v im M(-3; 1). Gi T1 v T2 l cc tip im ca cc tip tuyn k t M n(C). Vit phng trnh ng thng T1T2

2. Cho tp hp A gm n phn t (n 4). Bit rng s tp con gm 4 phn t caA bng 20 ln s tp con gm 2 phn t ca A. Tm k {1, 2,..., n} sao cho s tpcon gm k phn t ca A l ln nht.Cu5b:Theo chng trnh phn ban: (2 im)

1. Gii bt phng trnh: ( ) ( )25 5 5log 4 144 4log 2 1 log 2 1x x+ < + +

2. Cho hnh chp S.ABCD c y ABCD l hnh ch nht vi AB = a, AD =

a 2 , SA = a v SA vung gc vi mt phng (ABCD). Gi M v N ln lt l trungim ca AD v SC; I l giao im ca BM v AC. Chng minh rng: mt phng(SAC) vung gc vi mt phng (SMB). Tnh th tch ca khi t din ANIB


Cu1: (2 im)

Cho hm s y = x3 - 3x + 21. Kho st s bin thin v v th (C) ca hm s cho.2. Gi d l ng thng i qua im A(3; 2) v c h s gc l m. Tm m

ng thng d ct th (C) ti ba im phn bit.Cu2: (2 im)

1. Gii phng trnh: cos3x + cos2x - cosx - 1 = 0


18/155



2. Gii phng trnh: 22 1 3 1 0 x x x + + = (x R)Cu3: (2 im)

Trong khng gian vi h to Oxyz, cho im A(1; 2; 3) v hai ng thng

d1:2 2 3

2 1 1

x y z + = =

d2:1 1 1

1 2 1

x y z += =

1. Tm to im A i xng vi im A qua ng thng d12. Vit phng trnh ng thng i qua A vung gc vi d1 v ct d2

Cu4: (2 im)

1. Tnh tch phn: I = ( )1 20

2x

x e dx

2. Chng minh rng: vi mi a > 0, h phng trnh sau c nghim duy nht:( ) ( )ln 1 ln 1x ye e x y

y x a

= + +

=

Phn T chn: Th sinh chn Cu 5.a hc Cu 5.b

Cu5a:Theo chng trnh khng phn ban: (2 im)1. Trong mt phng vi h to Oxy cho ng trn (C): x2 + y2 - 2x - 2y + 1

= 0 v ng thng d: x - y + 3 = 0. Tm to im M nm trn d sao cho ngtrn tm M, c bn knh gp i bn knh ng trn (C) tip xc ngoi vi ngtrn (C)

2. i thanh nin xung kch ca mt trng ph thng c 12 hc sinh, gm 5hc sinh lp A, 4 hc sinh lp B v 3 hc sinh lp C. Cn chn 4 hc sinh i lm

nhim v, sao cho 4 hc sinh ny thuc khng qu 2 trong 3 lp trn. Hi c baonhiu cch chn nh vy?Cu5b:Theo chng trnh phn ban: (2 im)

1. Gii phng trnh:2 2 22 4.2 2 4 0 x x x x x+ + =

2. Cho hnh chp S.ABC c y ABC l tam gic u cnh a, SA = 2a v SAvung gc vi mt phng (ABC). Gi M v N ln lt l hnh chiu vung gc ca Atrn cc ng thng SB v SC. Tnh th tch ca khi chp A.BCNM

s 18

Phn chung c tt c cc th sinhCu1: (2 im)

Cho hm s: y =( )2 22 1 4

2

x m x m m

x

+ + + ++

(1) m l tham s

1. Kho st s bin thin v v th ca hm s (1) khi m = -1.2. Tm m hm s (1) c cc i v cc tiu, ng thi cc im cc tr ca

th cng vi gc to to thnh mt tam gic vung ti O


19/155



Cu2: (2 im)

1. Gii phng trnh: ( ) ( )2 21 sin cos 1 cos sin 1 sin 2 x x x x x+ + + = + 2. Tm m phng trnh sau c nghim thc: 243 1 1 2 1 x m x x + + =

Cu3: (2 im)Trong khng gian vi h to Oxyz cho hai ng thng

d1:1 2

2 1 1

x y z += =

v d2:

1 2

1

3

x t

y t

z

= + = + =

1. Chng minh rng: d1 v d2 cho nhau.2. Vit phng trnh ng thng d vung gc vi mt phng (P): 7x + y - 4z =

0 v ct hai ng thng d1, d2Cu4: (2 im)

1. Tnh din tch hnh phng gii hn bi cc ng: y = (e + 1)x, y = (1 + ex)x2. Cho x, y, z l cc s thc dng thay i v tho mn iu kin: xyz = 1.

Tm GTNN ca biu thc: P =( ) ( ) ( )2 2 2

2 2 2

x y z y z x z x y

y y z z z z x x x x y y

+ + ++ +

+ + +


Cu5a:Theo chng trnh khng phn ban: (2 im)

1. Trong mt phng vi h to Oxy cho ABC c A(0; 2) B(-2 -2) vC(4; -2). Gi H l chn ng cao k t B; M v N ln lt l trung im ca cc cnhAB v BC. Vit phng trnh ng trn i qua cc im H, M, N

2. Chng minh rng:

21 3 5 2 1

2 2 2 2

1 1 1 1 2 1

...2 4 6 2 2 1

nn

n n n nC C C C n n

+ + + + = + Cu5b:Theo chng trnh phn ban: (2 im)

1. Gii bt phng trnh: ( ) ( )3 13

2log 4 3 log 2 3 2x x + +

2. Cho hnh chp S.ABCD c y l hnh vung cnh a, mt bn SAD l tamgic u v nm trong mt phng vung gc vi y. Gi M, N, P ln lt l trungim ca cc cnh SB, BC, CD. Chng minh AM vung gc vi BP v tnh th tchca khi t din CMNP.


Cu1: (2 im)Cho hm s: y = -x3 + 3x2 + 3(m2 -1)x - 3m2 - 1 (1) m l tham s1. Kho st s bin thin v v th ca hm s (1) khi m = 12. Tm m hm s (1) c cc i, cc tiu v cc im cc tr ca th hm

s (1) cch u gc to O.Cu2: (2 im)


20/155



1. Gii phng trnh: 2sin22x + sin7x - 1 = sinx2. Chng minh rng vi mi gi tr dng ca tham s m, phng trnh sau c

hai nghim thc phn bit: x2 + 2x - 8 = ( )2m x

Cu3: (2 im)Trong khng gian vi h to Oxyz cho mt cu (S): x2 + y2 + z2 - 2x + 4y +

2z - 3 = 0 v mt phng (P): 2x - y + 2z - 14 = 01. Vit phng trnh mt phng (Q) cha trc Ox v ct (S) theo mt ng

trn c bn knh bng 3.2. Tm to im M thuc mt cu (S) sao cho khong cch t M n mt

phng (P) ln nhtCu4: (2 im)

1. Cho hnh phng H gii hn bi cc ng: y = xlnx, y = 0, x = e. Tnh thtch ca khi trn xoay to thnh khi quay hnh H quanh trc Ox.

2. Cho x, y, z l ba s thc dng thay i. Tm gi tr nh nht ca biu thc:P =

1 1 1

2 2 2

x y z x y z

yz zx xy

+ + + + +


Cu5a:Theo chng trnh khng phn ban: (2 im)1. Tm h s ca s hng cha x10 trong khai trin nh thc ca (2 + x)n bit

( )0 1 1 2 2 3 33 3 3 3 ... 1 2048nn n n n n

n n n n nC C C C C + + + =

2. Trong mt phng vi h to Oxy cho im A(2; 2) v cc ng thng:d1: x + y - 2 = 0 d2: x + y - 8 = 0

Tm to cc im B v C ln lt thuc d1 v d2 sao cho ABC vung cn tiA.Cu5b:Theo chng trnh phn ban: (2 im)

1. Gii phng trnh: ( ) ( )2 1 2 1 2 2 0x x

+ =

2. Cho hnh chp t gic u S.ABCD c y l hnh vung cnh a. Gi E lim i xng ca D qua trung im ca SA, M l trung im ca AE, N l trungim ca BC. Chng minh MN vung gc vi BD v tnh theo a khong cch gia haing thng MN v AC.


Cu1: (2 im) Cho hm s: y =2

1

x

x +

1. Kho st s bin thin v v th (C) ca hm s cho.2. Tm to im M thuc (C), bit tip tuyn ca (C) ti M ct hai trc Ox,

Oy ti A, B v tam gic OAB c din tch bng1

4


21/155



Cu2: (2 im)

1. Gii phng trnh: 2sin cos 3 cos 22 2

x xx

+ + =

2. Tm gi tr ca tham s m h phng trnh sau c nghim thc:

3 3

3 3

1 15

1 115 10

x y

x y

x y mx y

+ + + = + + + =

Cu3: (2 im)Trong khng gian vi h to Oxyz cho hai im A(1; 4; 2 B(-1 2; 4) v

ng thng :1 2

1 1 2

x y z += =

1. Vit phng trnh ng thng d i qua trng tm G ca tam gic OAB vvung gc vi mt phng (OAB).

2. Tm to im M thuc ng thng sao cho MA2 + MB2- nh nhtCu4: (2 im)

1. Tnh tch phn: I = 3 21

ln

e

x xdx

2. Cho a b > 0. Chng minh rng: 1 12 22 2

b a

a b

a b

+ +

Phn T chn: Th sinh chn Cu 5.a hc Cu 5.bCu5a:Theo chng trnh khng phn ban: (2 im)

1. Tm h s ca x5 trong khai trin thnh a thc ca: x(1 - 2x)5 + x2(1 + 3x)102. Trong mt phng vi h to Oxy cho ng trn (C): (x - 1)2 + (y + 2)2 =

9 v ng thng d: 3x - 4y + m = 0.

Tm m trn d c duy nht mt im P m t c th k c hai tip tuynPA, PB ti (C) (A, B l cc tip im) sao cho PAB uCu5b:Theo chng trnh phn ban: (2 im)

1. Gii phng trnh: ( )2 2 1log 4 15.2 27 2log 04.2 3

x x

x+ + + =

2. Cho hnh chp S.ABCD c y l hnh thang, ABC = BAD = 900 , BA =BC = a, AD = 2a. cnh bn SA vung gc vi y v SA = a 2 . Gi H lhnh chiu vung gc ca A trn SB. Chng minh tam gic SCD vung vtnh theo a khong cch t H n mt phng (SCD)

s 21Cu1: (2 im)

Cho hm s: y = x4 - mx2 + m - 1 (1) (m l tham s)

1) Kho st s bin thin v v th ca hm s (1) khi m = 8.2) Xc nh m sao cho th ca hm s (1) ct trc honh ti bn im phn bit.

Cu2: (2 im)


22/155



1) Gii bt phng trnh: ( ) ( )xxx 2.32log44log 122

1

2

1 ++

2) Xc nh m phng trnh: ( ) 02sin24coscossin4 44 =+++ mxxxx c t

nht mt nghim thuc on

2

;0

Cu3: (2 im)

1) Cho hnh chp S.ABC c y ABC l tam gic u cnh a v cnh bn SA vunggc vi mt phng y (ABC). Tnh khong cch t im A ti mt phng (SBC) theo

a, bit rng SA =2

6a


1

02

3

1x

dxx

Cu4: (2 im)Trong mt phng vi h to cc vung gc Oxy, cho hai ng trn:

(C1): x2 + y2 - 10x = 0, (C2): x

2 + y2 + 4x - 2y - 20 = 01) Vit phng trnh ng trn i qua cc giao im ca (C1), (C2) v c tm nm

trn ng thng x + 6y - 6 = 0.2) Vit phng trnh tip tuyn chung ca cc ng trn (C1) v (C2).

Cu5: (2 im)

1) Gii phng trnh: 16212244 2 +=++ xxxx

2) i tuyn hc sinh gii ca mt trng gm 18 em, trong c 7 hc sinh khi12, 6 hc sinh khi 11 v 5 hc sinh khi 10. Hi c bao nhiu cch c 8 hc sinhtrong i i d tri h sao cho mi khi c t nht mt em c chn.

Cu6: ( Tham kho)

Gi x, y, z l khong cch t im M thuc min trong ca ABC c 3 gc nhn n

cc cnh BC, CA, AB. Chng minh rng:R

cbazyx

2

222 ++++ ; a, b, c l

ba cnh ca , R l bn knh ng trn ngoi tip. Du "=" xy ra khi no?

s 22

Cu1: (2 im)


23/155



1) Tm s n nguyn dng tho mn bt phng trnh: nCA nnn 9223 + , trong

knA v

knC ln lt l s chnh hp v s t hp chp k ca n phn t.

2) Gii phng trnh: ( ) ( ) ( )xxx 4log1log4

13log

2

12

842 =++

Cu2: (2,5 im)

Cho hm s: y =2

22

+

x

mxx(1) (m l tham s)

1) Xc nh m hm s (1) nghch bin trn on [-1; 0].


3) Tm a phng trnh sau c nghim:

( ) 01232922

1111 =+++ ++ aa tt

Cu3: (1,5 im)

1) Gii phng trnh:x

xgx

xx

2sin8

12cot

2

1

2sin5

cossin44

=+

2) Xt ABC c di cc cnh AB = c; BC = a; CA = b. Tnh din tch ABC, bit

rng: bsinC(b.cosC + c.cosB) = 20

Cu4: (3 im)

1) Cho t din OABC c ba cnh OA; OB v OC i mt vung gc. Gi ; ; ln

lt l cc gc gia mt phng (ABC) vi cc mt phng (OBC); (OCA) v (OAB).

Chng minh rng: 3coscoscos ++ .

2) Trong khng gian vi h to cc Oxyz cho mt phng (P): x- y + z + 3 = 0

v hai im A(-1; -3; -2), B(-5; 7; 12).

a) Tm to im A' l im i xng vi im A qua mt phng (P).

b) Gi s M l mt im chy trn mt phng (P), tm gi tr nh nht ca biu thc:

MA + MB.

Cu5: (1,0 im)

Tnh tch phn: I =

( )

+

3ln

03

1x

x

e

dxe

s 23

Cu1: (3,0 im)


24/155



Cho hm s: y =3

122

3

1 23 + mxmxx (1) (m l tham s)

1) Cho m =2

1

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

b) Vit phng trnh tip tuyn ca th (C), bit rng tip tuyn song song ving thng d: y = 4x + 2.

2) Tm m thuc khong

6

5;0 sao cho hnh phng gii hn bi th ca hm s (1)

v cc ng x = 0, x = 2, y = 0 c din tch bng 4.Cu2: (2 im)

1) Gii h phng trnh:

=

=+

0loglog

034

24 yx

yx

2) Gii phng trnh:x

xxxtg4

24

cos

3sin2sin21 =+

Cu3: (2 im)1) Cho hnh chp S.ABCD c y ABCD l hnh vung cnh a, SA vung gc vi

mt phng (ABCD) v SA = a. Gi E l trung im ca cnh CD. Tnh theo a khongcch t im S n ng thng BE.2) Trong khng gian vi h to cc Oxyz cho ng thng

:

=+++

=+++

02

012

zyx

zyxv mt phng (P): 4x - 2y + z - 1 = 0

Vit phng trnh hnh chiu vung gc ca ng thng trn mt phng (P).Cu4: (2 im)

1) Tm gii hn: L =x

xx

x

3

0

11lim

++

2) Trong mt phng vi h ta cac Oxy cho hai ng trn:(C1): x

2 + y2 - 4y - 5 = 0 v (C2): x2 + y2 - 6x + 8y + 16 = 0

Vit phng trnh cc tip tuyn chung hai ng trn (C1) v (C2)

Cu5: (1 im)Gi s x, y l hai s dng thay i tho mn iu kin x + y =

4

5. Tm gi tr nh

nht ca biu thc: S =yx 4

14 +

s 24


25/155



Cu1: (2 im)

1) Gii bt phng trnh: 12312 +++ xxx

2) Gii phng trnh: tgx + cosx - cos2x = sinx(1 + tgxtg2

x)

Cu2: (2 im)Cho hm s: y = (x - m)3 - 3x (m l tham s)

1) Xc nh m hm s cho t cc tiu ti im c honh x = 0.

2) Kho st s bin thin v v th ca hm s cho khi m = 1.

3) Tm k h bt phng trnh sau c nghim:( )

+


26/155



Cu1: (2 im)

Cho hm s: y =x

mxx

+

1

2

(1) (m l tham s)

1) Kho st s bin thin v v th ca hm s (1) khi m = 0.2) Tm m hm s (1) c cc i v cc tiu. Vi gi tr no ca m th khong cch

gia hai im cc tr ca th hm s (1) bng 10.Cu2: (2 im)

1) Gii phng trnh: 0log3log16 2327 3 = xx xx

2) Cho phng trnh: axx

xx =+++

3cos2sin

1cossin2(2) (a l tham s)

a) Gii phng trnh (2) khi a =3

1.

b) Tm a phng trnh (2) c nghim.Cu3: (3 im)1) Trong mt phng vi h ta cac Oxy cho ng thng d: x - y + 1 = 0 v

ng trn (C): x2 + y2 + 2x - 4y = 0. Tm to im M thuc ng thng d m qua ta k c hai ng thng tip xc vi ng trn (C) ti A v B sao cho gcAMB bng 600.2) Trong khng gian vi h to cc Oxyz cho ng thng

d:

=+=+

0422

0122

zyx

zyxv mt cu (S): x2 + y2 + z2 + 4x - 6y + m = 0.

Tm m ng thng d ct mt cu (S) ti hai im M, N sao cho khong cch giahai im bng 9.3) Tnh th tch khi t din ABCD, bit AB = a; AC = b; AD = c v cc gc BAC;

CAD; DAB u bng 600Cu4: (2 im)


0

56 3cossincos1

xdxxx

2) Tm gii hn:x

xx

x cos1

1213lim

23 2

0 ++

Cu5: (1 im)Gi s a, b, c, d l bn s nguyn thay i tho mn 1 a < b < c < d 50. Chng

minh bt ng thc:b

bb

d

c

b

a

50

502 +++ v tm gi tr nh nht ca biu thc:

S =d

c

d

a +


27/155



s 26

Cu1: (2 im)

1) Kho st s bin thin v v th ca hm s: y = xxx 323

1 23 +

2) Tnh din tch hnh phng gii hn bi th hm s (1) v trc honh.Cu2: (2 im)

1) Gii phng trnh: xx

sincos8

12

=

2) Gii h phng trnh:( )( )

=+

=+

3532log

3532log

23

23

xyyy

yxxx

y

x

Cu3: (2 im)

1) Cho hnh t din u ABCD, cnh a = 6 2 cm. Hy xc nh v tnh di on

vung gc chung ca hai ng thng AD v BC.

2) Trong mt phng vi h ta cac Oxy cho elip (E): 149

22

=+ yx v ng

thng dm: mx - y - 1 = 0.

a) Chng minh rng vi mi gi tr ca m, ng thng dm lun ct elp (E) ti hai

im phn bit.

b) Vit phng trnh tip tuyn ca (E), bit rng tip tuyn i qua im N(1; -3)

Cu4: (1 im)

Gi a1, a2, ..., a11 l h s trong khai trin sau:

( ) ( ) 119

210

11110

...21 axaxaxxx ++++=++

Hy tnh h s a5

Cu5: (2 im)

1) Tm gii hn: L = ( )2

6

1 1

56

lim +

x

xx

x

2) Cho ABC c din tch bng2

3. Gi a, b, c ln lt l di ca cc cnh BC,

CA, AB v ha, hb, hc tng ng l di cc ng cao k t cc nh A, B, C ca tam

gic. Chng minh rng: 3111111

++

++

cba hhhcba


28/155



s 27

Cu1: (2 im)

1) Kho st s bin thin v v th ca hm s y =( )12

3422

x

xx

2) Tm m phng trnh: 2x2 - 4x - 3 + 2m 1x = 0 c hai nghim phn bit.

Cu2: (2 im)

1) Gii phng trnh: ( ) 0623 =++ xcosxsintgxtgx

2) Gii h phng trnh:

=+

=

322yx

xy ylogxylog

Cu3: (3 im)

1) Trong mt phng vi h ta cc Oxy cho parabol (P) c phng trnh y2 = x

v im I(0; 2). Tm to hai im M, N thuc (P) sao cho INIM 4= .

2) Trong khng gian vi h to cc Oxyz cho t din ABCD vi A(2; 3; 2),

B(6; -1; -2), C(-1; -4; 3), D(1; 6; -5). Tnh gc gia hai ng thng AB v CD. Tm

to im M thuc ng thng CD sao cho ABM c chu vi nh nht.

3) Cho lng tr ng ABC. A'B'C' c y ABC l tam gic cn vi AB = AC = a v

gc BAC = 1200

, cnh bn BB' = a. Gi I l trung im CC'. Chng minh rng AB'Ivung A. Tnh cosin ca gc gia hai mt phng (ABC) v (AB'I).

Cu4: (2 im)

1) C bao nhiu s t nhin chia ht cho 5 m mi s c 4 ch s khc nhau?


4

02cos1

dxx

x

Cu5: (1 im)Tm gi tr ln nht v gi tr nh nht ca hm s: y = sin5x + 3 cosx

]


29/155



s 28

Cu1: (2 im)

Cho hm s: y =( )

( )mx

mmxmx

+

+++++

2

41222

(1) (m l tham s)

1) Tm m hm s (1) c cc tr v tnh khong cch gia hai im cc tr ca

th hm s (1).

2) Kho st s bin thin v v th ca hm s (1) khi m = 0

Cu2: (2 im)

1) Gii phng trnh: cos2x + cosx(2tg2x - 1) = 2

2) Gii bt phng trnh: 11 21212.15 ++ ++ xxx

Cu3: (3 im)1) Cho t din ABCD vi AB = AC = a, BC = b. Hai mt phng (BCD) v (ABC)

vung gc vi nhau v gc BDC = 900. Xc nh tm v tnh bn knh mt cu ngoi

tip t din ABCD thao a v b.

2) Trong khng gian vi h to cc Oxyz cho hai ng thng:

d1:12

1

1

zyx =+= v d2:

=+=+

012

013

yx

zx

a) Chng minh rng d1, d2 cho nhau v vung gc vi nhau.

b) Vit phng trnh tng qut ca ng thng d ct c hai ng thng d1, d2 v

song song vi ng thng :2

3

4

7

1

4

== zyx

Cu4: (2 im)

1) T cc ch s 0, 1, 2, 3, 4, 5 c th lp c bao nhiu s t nhin m mi s c 6

ch s khc nhau v ch s 2 ng cnh ch s 3?


0

231 dxxx

Cu5: (1 im)

Tnh cc gc ca ABC bit rng:( )

=

8

332

2sin

2sin

2sin

4

CBA

bcapp

trong BC = a, CA = b, AB = c, p =2

cba ++


30/155



s 29

Cu1: (2 im)

Cho hm s: y = (x - 1)(x2

+ mx + m) (1) (m l tham s)1) Tm m th hm s (1) ct trc honh ti ba im phn bit.


Cu2: (2 im)

1) Gii phng trnh: 032943 26 =++ xcosxcosxcos

2) Tm m phng trnh: ( ) 042

12

2 =+ mxlogxlog c nghim thuc

khong (0; 1).

Cu3: (3 im)

1) Trong mt phng vi h ta cc Oxy cho ng thng d: x - 7y + 10 = 0.

Vit phng trnh ng trn c tm thuc ng thng : 2x + y = 0 v tip xc vi

ng thng d ti im A(4; 2).

2) Cho hnh lp phng ABCD.A'B'C'D'. Tm im M thuc cnh AA' sao cho

mt phng (BD'M) ct hnh lp phng theo mt thit din c din tch nh nht.

3) Trong khng gian vi h to cc Oxyz cho t din OABC vi A(0; 0;

3a ), B(0; 0; 0), C(0; a 3 ; 0) (a > 0). Gi M l trung im ca BC. Tnh khong

cch gia hai ng thng AB v OM.

Cu4: (2 im)

1) Tm gi tr ln nht v gi tr nh nht ca hm s: y = x6 + ( )3214 x trnon [-1; 1].

2) Tnh tch phn: I =

5

2

2

1

ln

lnx

x

e

dxe

Cu5: (1 im)


31/155



T cc ch s 1, 2, 3, 4, 5, 6 c th lp c bao nhiu s t nhin, mi s c 6

ch s v tho mn iu kin: Su ch s ca mi s l khc nhau v trong mi s

tng ca ba ch s u nh hn tng ca ba ch s cui mt n v?

s 30

Cu1: (2 im)

Cho hm s: y =1

12

x

x(1)

1) Kho st s bin thin v v th ca hm s (C) ca hm s (1).2) Gi I l giao im ca hai ng tim cn ca (C). Tm im M thuc (C) sao cho

tip tuyn ca (C) ti M vung gc vi ng thng IM.

Cu2: (2 im)

1) Gii phng trnh:( )

11cos2

42sin2cos32

2

=

x

xx

2) Gii bt phng trnh: ( ) 06log1log2log 24

1

2

1 ++ xx

Cu3: (3 im)

1) Trong mt phng vi h ta cc Oxy cho elip (E): 114

22

=+ yx , M(-2; 3),

N(5; n). Vit phng trnh cc ng thng d1, d2 qua M v tip xc vi (E). Tm n trong s cc tip tuyn ca (E) i qua N v c mt tip tuyn song song vi d1 hoc d2

2) Cho hnh chp u S.ABC, y ABC c cnh bng a, mt bn to vi y mt gc

bng (00 < < 900). Tnh th tch khi chp S.ABC v khong cch t nh A nmt phng (SBC).3) Trong khng gian vi h to cc Oxyz cho hai im I(0; 0; 1), K(3; 0; 0).

Vit phng trnh mt phng i qua hai im I, K v to vi vi mt phng xOy mtgc bng 300

Cu4: (2 im)1) T mt t gm 7 hc sinh n v 5 hc sinh nam cn chn ra 6 em trong s hc

sinh n phi nh hn 4. Hi c bao nhiu cch chn nh vy?

2) Cho hm s f(x) =( )

xbxe

x

a ++ 31

. Tm a v b bit rng


32/155



f'(0) = -22 v ( ) 51

0

= dxxf

Cu5: (1 im)

Chng minh rng:2

2cos2

xxxex ++ x R

s 31

Cu1: (2 im)

Cho hm s: y =3

6522

++++

xmxx

(1) (m l tham s)


2) Tm m hm s (1) ng bin trn khong (1; + ).Cu2: (2 im)

1) Gii phng trnh:( )

( )xsinxcosxsin

xcosxcos +=+

12

12

2) Cho hm s: f(x) = 2xlogx (x > 0, x 1)

Tnh f'(x) v gii bt phng trnh f'(x) 0

Cu3: (3 im)

1) Trong mt phng vi h ta cc Oxy cho ABC c nh A(1; 0) v hai

ng thng ln lt cha cc ng cao v t B v C c phng trnh tng ng l:

x - 2y + 1 = 0 v 3x + y - 1 = 0 Tnh din tch ABC.

2) Trong khng gian vi h to cc Oxyz cho mt phng

(P): 2x + 2y + z - m2 - 3m = 0 (m l tham s)

v mt cu (S): ( ) ( ) ( ) 9111 222 =+++ zyx

Tm m mt phng (P) tip xc vi mt cu (S). Vi m tm c, hy xc nh to

tip im ca mt phng (P) v mt cu (S).

3) Cho hnh chp S.ABC c y ABC l tam gic vung ti B, AB = a, BC = 2a, cnh

SA vung gc vi y v SA = 2a. Gi M l trung im ca SC. Chng minh rng

AMB cn ti M v tnh din tch AMB theo a.Cu4: (2 im)

1) T 9 ch s 0, 1, 2, 3, 4, 5, 6, 7, 8 c th lp c bao nhiu s t nhin chn m

mi s gm 7 ch s khc nhau?


33/155




0

32

dxex x

Cu5: (1 im)

Tm cc gc A, B, C ca ABC biu thc: Q = CsinBsinAsin 222 + t gi tr

nh nht.

s 32

Cu1: (2 im)

1) Kho st s bin thin v v th ca hm s (C) ca hm s: y = 2x3 - 3x2 - 1

2) Gi dk l ng thng i qua im M(0 ; -1) v c h s gc bng k. Tm k

ng thng dk ct (C) ti ba im phn bit.

Cu2: (2 im)

1) Gii phng trnh:xsinxcos

tgxgxcot2

42+=

2) Gii phng trnh: ( ) xlog x = 1455 Cu3: (3 im)

1) Trong khng gian vi h to cc Oxyz cho hai im A(2; 1; 1), B(0; -1; 3)

v ng thng d:

=+

=

083

01123

zy

yx

a) Vit phng trnh mt phng (P) i qua trung im I ca AB v vung gc vi

AB. Gi K l giao im ca ng thng d v mt phng (P), chng minh rng d

vung gc vi IK.

b) Vit phng trnh tng qut ca hnh chiu vung gc ca d trn mt phng c

phng trnh: x + y - z + 1 = 0.

2) Cho t din ABCD c AD vung gc vi mt phng (ABC) v ABC vung ti A,

AD = a, AC = b, AB = c. Tnh din tch ca BCD theo a, b, c v chng minh rng:

2S ( )cbaabc ++

Cu4: (2 im)

1) Tm s t nhin n tho mn: 1002 333222 =++ nnnnnnnn CCCCCC

trong knC l s t hp chp k ca n phn t.


34/155



2) Tnh tch phn: I = +e

xdxlnx

x

1

21

Cu5: (1 im)

Xc nh dng ca ABC, bit rng: ( ) ( ) BsinAsincBsinbpAsinap =+ 22

trong BC = a, CA = b, AB = c, p =2

cba ++

s 33

Cu1: (2,5 im)

1) Cho hm s: y =1

12

+

xmxx

(*)

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

b) Tm nhng im trn (C) c to l nhng s nguyn.

c) Xc nh m ng thng y = m ct th ca hm s (*) ti hai im phn bit

A, B sao cho OA vung gc vi OB.

Cu2: (1 im)

Cho ng trn (C): x2 + y2 = 9 v im A(1; 2). Hy lp phng trnh ca ng

thng cha dy cung ca (C) i qua A sao cho di dy cung ngn nht.

Cu3: (3,5 im)

1) Cho h phng trnh:

+=+=+

12

3

mymx

myx

a) Gii v bin lun h phng trnh cho.

b) Trong trng hp h c nghim duy nht, hy tm nhng gi tr ca m sao

cho nghim (x0; y0) tho mn iu kin

>>

0

0

0

0

y

x

2) Gii cc phng trnh v bt phng trnh sau:

a) sin(cosx) = 1b) 11252 5


35/155



b) 6 ng trn phn bit.

2) T kt qu ca 1) hy suy ra s giao im ti a ca tp hp cc ng ni trn.

Cu5: (2 im)

Cho hnh chp t gic u S.ABCD c cc cnh bn bng a v mt cho SAC l tam

gic u.

1) Tm tm v bn knh ca mt cu ngoi tip hnh chp.

2) Qua A dng mt phng () vung gc vi SC. Tnh din tch thit din to bi

mt phng () v hnh chp.

s 34

Cu1: (2 im)

Cho hm s: y =12

1

xx

1) Kho st s bin thin v v th ca hm s.

2) Tm cc im trn th hm s c to l cc s nguyn.

Cu2: (2 im)

1) Gii phng trnh: xsinxcostgxxtg 33

12 =

2) Gii bt phng trnh: ( ) ( ) ( ) 04221 33

1

3

1


36/155



Cu5: (2 im)

1) Cho a gic li c n cnh. Xc nh n a gic c s ng cho gp i s

cnh.

2) Tnh tch phn: I =( ) ++

1

0

2

11dx

xx

x

s 35

Cu1: (3,5 im)

Cho hm s: y = 14

2

+

x

xx

(1)

1) Kho st s bin thin v v th (C) ca hm s (1).

2) Tm m ng thng (d) c phng trnh y = mx ct (C) ti hai im phn bit.

3) Tnh din tch hnh phng c gii hn bi (C); tim cn xin v cc ng

thng x = 2; x = 4.

Cu2: (1 im)

Gii phng trnh: ( ) ( ) 021223 =++++ xcosxsinxsinxcosxsin

Cu3: (2 im)

Cho phng trnh: 04 22 =+ xx (2)


2) Xc nh m phng trnh (2) c nghim.

Cu4: (1 im)

Cho cc ch s: 0, 1, 2, 3, 4. C bao nhiu s t nhin chn gm 5 ch s khc nhau

lp t cc ch s trn?Cu5: ( 2,5 im)

Cho elip (E) c hai tiu im l F1( 03; ); ( )032 ;F v mt ng chun c

phng trnh: x =3

4.


37/155



1) Vit phng trnh chnh tc ca (E).

2) M l im thuc (E). Tnh gi tr ca biu thc:

P = MF.MFOMMFMF 2122

22

1 3 +

3) Vit phng trnh ng thng (d) song song vi trc honh v ct (E) ti hai

im A, B sao cho OA OB.

s 36

Cu1: (2,5 im)

Cho hm s: y =x

xx 232 +

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

2) Tm trn ng thng x = 1 nhng im M sao cho t M k c hai tip tuyn

ti (C) v hai tip tuyn vung gc vi nhau.

Cu2: (1,5 im) Gii cc phng trnh:

1) ( ) ( ) 24224 =+ xloglogxloglog

2)5

5

3

3 xsinxsin =

Cu3: (2 im)

Gii cc bt phng trnh:

1) ( ) ( ) 06140252 1 + xxx

Cu4: (2 im) Cho In = ( ) 1

0

221 dxxx

nv J n = ( )

1

0

21 dxxx

n

vi n nguyn dng.

1) Tnh Jn v chng minh bt ng thc:( )12

1

+

nIn

2) Tnh In + 1 theo In v tmn

n

x II

lim 1+

Cu5: (2 im)


38/155



1) Trong mt phng (P) cho ng thng (D) c nh, A l mt im c nh nm

trn (P) v khng thuc ng thng (D); mt gc vung xAy quay quanh A, hai tia

Ax v Ay ln lt ct (D) ti B v C. Trn ng thng (L) qua A v vung gc vi

(P) ly im S c nh khc A. t SA = h v d l khong cch t im A n (D).

Tm gi tr nh nht ca th tch t din SABC khi xAy quay quanh A.

2) Trong mt phng vi h ta cc Oxy cho ABC. im M(-1; 1) l trungim ca cnh BC; hai cnh AB v AC theo th t nm trn hai ng thng c

phng trnh l: x + y - 2 = 0; 2x + 6y + 3 = 0.

Xc nh to ba nh A, B, C.

s 37

Cu1: (3 im)

Cho hm s: y = x3 - 3mx + 2 c th l (Cm) (m l tham s)

1) Kho st s bin thin v v th (C1) ca hm s khi m = 1.

2) Tnh din tch hnh phng gii hn bi (C1) v trc honh.

3) Xc nh m (Cm) tng ng ch c mt im chung vi trc honh.

Cu2: (1 im)

1) Chng minh rng vi mi s nguyn dng n ta u c:

nnnnn

nnnnn C...CCCC...CCC

22

42

22

02

122

52

32

12 ++++=++++

2) T cc ch s 1, 2, 3, 4, 5 c th lp c bao nhiu s gm 3 ch s khc nhau

nh hn 245.

Cu3: (1,5 im)

1) Gii h phng trnh:( )( )( )( )

=++

=

15

3

22

22

yxyx

yxyx

2) Gii phng trnh: xx +=+ 173

Cu4: (1,5 im)Cho phng trnh: ( ) 01122 =++ mxcosmxcos (m l tham s)

1) Gii phng trnh vi m = 1.

2) Xc nh m phng trnh c nghim trong khong

;

2.

Cu5: (3 im)


39/155



1) Cho khi chp t gic u S.ABCD c cc cnh bn v cnh y u bng a. Gi

M, N v P ln lt l trung im ca cc cnh AD, BC v SC. Mt phng (MNP) ct

SD ti Q. Chng minh rng MNPQ l hnh thang cn v tnh din tch ca n.


(D1):

== =

tz

ty tx1

v (D2):

===

'tz

'ty 'tx 12

(t, t' R)

a) Chng minh (D1), (D2) cho nhau v tnh khong cch gia hai ng thng y.

b) Tm hai im A, B ln lt trn (D1), (D2) sao cho AB l on vung gc chung

ca (D1) v (D2).

s 38

Cu1: (3 im)

Cho hm s: y =1

12

+

xmxx

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

2) Xc nh m hm s ng bin trn cc khong (- ; 1) v (1; + )3) Vi gi tr no ca m th tim cn xin ca th hm s to vi cc trc to

mt tam gic c din tch bng 4 (n v din tch).

Cu2: (2 im)

Cho phng trnh: ( ) ( ) mtgxtgx

=++ 223223

1) Gii phng trnh khi m = 6.

2) Xc nh m phng trnh c ng hai nghim phn bit nm trong khong

22; .

Cu3: (2 im)

1) Gii bt phng trnh: ( )4

3

16

1313

4

14

xx loglog

2) Tnh tch phn: I =

2

0

32 xdxsinxsinxsin


40/155



Cu4: (2 im)

Trong mt phng vi h ta cc Oxy cho ABC v im M(-1; 1) l trung

im ca AB. Hai cnh AC v BC theo th t nm trn hai ng:

2x + y - 2 = 0 v x + 3y - 3 = 0

1) Xc nh ta ba nh A, B, C ca tam gic v vit phng trnh ng cao CH.2) Tnh din tch ABC.

Cu5: (1 im)

Gi s x, y l cc nghim ca h phng trnh:

+=+

=+

32

12

222 aayx

ayx

Xc nh a tch P = x.y t gi tr nh nht.

s 39

Cu1: (2 im)

Cho hm s: y =2

52

+

xxx

1) Kho st s bin thin v v th ca hm s cho.

2) Bin lun theo m s nghim ca phng trnh:2

52

+

x

xx= m

Cu2: (2 im)

1) Gii phng trnh: 01 =++ xcosxsin

2) Gii bt phng trnh: ( ) xlogxlog x 22

22 + 4

Cu3: (1 im)

Gii h phng trnh:( )

++=+

=

2

7

22

33

yxyx

yxyx

Cu4: (1,5 im)

Tnh cc tch phn sau: I1 = ( )

+2

0

442 dxxcosxsinxcos I2 =

2

0

5 xdxcos

Cu5: (3,5 im)


41/155



1) Trong mt phng vi h ta cc Oxy cho ng trn (S) c phng trnh:

x2 + y2 - 2x - 6y + 6 = 0 v im M(2 ; 4)

a) Chng minh rng im M nm trong ng trn.

b) Vit phng trnh ng thng i qua im M, ct ng trn ti hai im A v B

sao cho M l trung im ca AB.c) Vit phng trnh ng trn i xng vi ng trn cho qua ng thng

AB.

2) Cho hnh chp t gic S.ABCD c di tt c cc cnh u bng a. Chng minh

rng:

a) y ABCD l hnh vung.

b) Chng minh rng nm im S, A, B, C, D cng nm trn mt mt cu. Tm tm v

bn knh ca mt cu .

s 40

Cu1: (2 im)

Cho hm s: y =( )

( )1

1322

++

mxmxmx


2) Tm tt c cc gi tr ca m hm s cho ng bin trong khong (0; + ).Cu2: (2 im)

1) Tnh tch phn: I = ( )

2

0

33 dxxsinxcos

2) T 5 ch s 0, 1, 2, 5, 9 c th lp c bao nhiu s l, mi s gm 4 ch s

khc nhau.

Cu3: (3 im)1) Gii phng trnh: ( ) 442 =+ xsinxcosxsin

2) Gii h phng trnh:

+=

+=

432

432

22

22

yxy

xyx

3) Cho bt phng trnh: ( ) ( ) 114 2525


42/155



Tm m bt phng trnh nghim ng vi mi x thuc khong (2 ; 3)

Cu4: (3 im)

Trong khng gian vi h to cc Oxyz cho hai ng thng (1) v (2) c

phng trnh: 1

:

=+

=+

0104

0238

zy

yx

2

:

=++

=

022

032

zy

zx

1) Chng minh (1) v (2) cho nhau.

2) Vit phng trnh ng thng () song song vi trc Oz v ct cc ng thng

(1) v (2).

s 41

Cu1: (2,5 im)

Cho hm s: y = x3 - mx2 + 1 (Cm)

1) Khi m = 3

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

b) Tm trn th hm s tt c cc cp im i xng nhau qua gc to .

2) Xc nh m ng cong (Cm) tip xc vi ng thng (D) c phng trnh

y = 5. Khi tm giao im cn li ca ng thng (D) vi ng cong (Cm).

Cu2: (1,5 im)

1) Gii bt phng trnh: ( ) ( ) 13

3

1

310310

++

+ xx

xx

0

2) Gii phng trnh: ( ) 01641 323 =++ xlogxxlogx

Cu3: (2 im)

1) Gii phng trnh: ( )( ) 45252 =++++ xxxx

2) Gii phng trnh:xcos

xcosxcos1

7822 =+

Cu4: (2 im)


43/155



1) Trong khng gian vi h to cc Oxyz cho im A(-1; 2; 5), B(11; -16; 10).

Tm trn mt phng Oxy im M sao cho tng cc khong cch t M n A v B l b

nht.


3

2 48

7

21

dxxx

x

Cu5: (2 im)

Trn tia Ox, Oy, Oz i mt vung gc ln lt ly cc im khc O l M, N v S

vi OM = m, ON = n v OS = a.

Cho a khng i, m v n thay i sao cho m + n = a.

1) a) Tnh th tch hnh chp S.OMN

b) Xc nh v tr ca cc im M v N sao cho th tch trn t gi tr ln nht.

2) Chng minh:

s 42

Cu1: (2 im)

1) Kho st s bin thin v v th (C) ca hm s: y =2

1

+

x

x

2) Tm cc im trn th (C) ca hm s c to l nhng s nguyn.

3) Tm cc im trn th (C) sao cho tng khong cch t im n hai tim

cn l nh nht.

Cu2: (2 im)

1) Gii phng trnh: 012315 = xxx

2) Gii h phng trnh:( )

( )

=+

=+

223

223

xylog

yxlog

y

x

Cu3: (1 im)

Gii phng trnh lng gic: 022 3 =+ xcosxcosxsin

Cu4: (2 im)


44/155



Cho D l min gii hn bi cc ng y = tg2x; y = 0; x = 0 v x =4

.

1) Tnh din tch min D.

2) Cho D quay quanh Ox, tnh th tch vt th trn xoay c to thnh.

Cu5: (1,5 im)Trong khng gian vi h to cc Oxyz cho ba im A(1; 4; 0), B(0; 2; 1),

C(1; 0; -4).

1) Vit phng trnh tng qut ca mt phng () i qua im C v vung gc vi

ng thng AB.

2) Tm to im C' i xng vi im C qua ng thng AB.

Cu6: (1,5 im)

1) Gii phng trnh: xxCCC xxx 14966 2321 =++ (x 3, x N)

2) Chng minh rng: 1919201720

520

320

120 2=+++++ CC...CCC

s 43

Cu1: (2,5 im)

1) Kho st s bin thin v v th ca hm s y = 1

2

xx .

2) Bin lun theo tham s m s nghim ca phng trnh: mxx = 1

2

Cu2: (2,5 im)

1) Chng minh rng nu x, y l hai s thc tho mn h thc:

x + y = 1 th x4 + y48

1

2) Gii phng trnh: 12822324222

212 ++>++ + x.x..xx xxx

Cu3: (2,5 im)


22

=+xcos

xcosxsinxsin


45/155



2) Cc gc ca ABC tho mn iu kin:

( )CcosBcosAcosCsinBsinAsin 222222 3 ++=++ Chng minh rng ABC l tam gic u.

Cu4: (2,5 im)

1) Tnh tch phn: e

xdxlnx1

22

2) Cho hnh lp phng ABCD.A'B'C'D' vi cc cnh bng a. Gi s M, N ln

lt l trung im ca BC, DD'. Tnh khong cch gia hai ng thng BD v MN

theo a.

s 44

Cu1: (3 im)

Cho hm s: y = x3 - 3mx2 + 3(2m - 1)x + 1 (1)1) Kho st s bin thin v v th ca hm s (1) khi m = 2.

2) Xc nh m sao cho hm s (1) ng bin trn tp xc nh.

3) Xc nh m sao cho hm s (1) c mt cc i v mt cc tiu. Tnh to ca

im cc tiu.

Cu2: (2 im)

1) Gii phng trnh: 23sin2sinsin 222 =++ xxx

2) Tm m phng trnh: ( )33 2422

122 =+ xlogmxlogxlog

c nghim thuc khong [32; + ).

Cu3: (2 im)


46/155



1) Gii h phng trnh:

=+

=+

015132

932

22

22

yxyx

yxyx

2) Tnh tch phn: e

dxx

xln

13

Cu4: (1,5 im)

Cho hnh chp S.ABC c y ABC l tam gic u cnh a v SA vung gc vi mt

phng (ABC). t SA = h.

1) Tnh khong cch t A n mt phng (SBC) theo a v h.

2) Gi O l tm ng trn ngoi tip tam gic ABC v H l trc tm tam gic SBC.

Chng minh: OH (SBC).

Cu5: (1,5 im)Trong khng gian vi h to cc Oxyz cho ng thng d v mt phng (P):

d:

==+032

03

zy

zx(P): x + y + z - 3 = 0

1) Vit phng trnh mt phng (Q) cha ng thng d v qua im M(1; 0; -2).

2) Vit phng trnh hnh chiu vung gc ca ng thng d trn mt phng (P).

s 45

Cu1: (3 im)

Cho hm s: y =1

12

xxx

(C)

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

2) Lp phng trnh tip tuyn vi (C) ti im c honh x = 0.

3) Tm h s gc ca ng thng ni im cc i, cc tiu ca th (C).

Cu2: (2,5 im)

1) Gii phng trnh: xxx .4269 =+ .

2) Tnh: ++

2

02

3

12

3

xx

dxx

Cu3: (2,5 im)


47/155



1) Gii h phng trnh:

=+

=+

26

2

33 yx

yx

2) Tnh gc C ca ABC nu: ( )( ) 211 =++ gBcotgAcot

Cu4: (2 im)

Trong khng gian vi h to cc Oxyz :

1) Cho 2 ng thng:

(1):

==

0

0

y

x(2):

==+

0

01

z

yx

Chng minh (1) v (2) cho nhau.

2) Cho 2 im A(1 ; 1 ; -1), B(3 ; 1 ; 1) v mt phng (P) c phng trnh:

x + y + z - 2 = 0Tm trn mt phng (P) cc im M sao cho MAB l tam gic u.

s 46Cu1: (2,5 im)

Cho hm s: y = x3 - (2m + 1)x2 - 9x (1)

1) Vi m = 1;

a) Kho st s bin thin v v th (C) ca hm s (1).

b) Cho im A(-2; -2), tm to im B i xng vi im A qua tm i xng

ca th (C).

2) Tm m th ca hm s (1) ct trc honh ti ba im phn bit c cc honh

lp thnh mt cp s cng.Cu2: (2 im)

1) Gii phng trnh: 03sin2cos4cossin =+ xxxx

2) Cho ABC cnh a, b, c tho mn h thc: 2b = a + c.

Chng minh rng: 32

cot2

cot =CgAg .


48/155



Cu3: (2 im)

1) Gii bt phng trnh: ( ) ( )12lg2

13lg

22 +> xxx

2) Tm a h phng trnh sau c nghim duy nht:( )

( )

=+

=+

1

1

2

2

xayxy

yaxxy

Cu4: (1,5 im)

1) Tnh tch phn: I = +++2

05cos3sin4

1sin3cos4

dxxx

xx

2) Tnh tng: P = 51054

1043

1032

1021

10110 33333 CCCCCC ++

1010

10910

9810

8710

7610

633333 CCCCC +++

Cu5: (2 im)1) Trong khng gian vi h to cc Oxyz cho mt phng (P) v mt cu (S) ln

lt c phng trnh: (P): y - 2z + 1 = 0 (S): x2 + y2 + z2 - 2z = 0.

Chng minh rng mt phng (P) v mt cu (S) ct nhau. Xc nh tm v bn knh

ca ng trn giao tuyn.

2) Cho hnh chp u S.ABC nh S, chiu cao l h, y l tam gic u cnh a. Qua

cnh AB dng mt phng vung gc vi SC. Tnh din tch thit din to thnh theo a

v h.

s 47

Cu1: (2,5 im)

Cho hm s: y =1

2222

+++

xmxmx

(m l tham s)


2) Tm m trn th c hai im i xng nhau qua gc to .

Cu2: (2 im)


122 =+ +++ xxxx .

2) Cho ABC. Chng minh rng nuCsin

BsintgCtgB

2

2

= th tam gic l tam gic vung

hoc cn.


49/155



Cu3: (2 im)

1) Tnh tch phn: 9

1

3 1 dxxx

2) Gii h phng trnh: ( )

+=+

+=+

yxyx

yyxx

322

22

Cu4: (2,5 im)

1) Cho hnh chp tam gic u S.ABC c gc gia mt bn v mt y l v SA =

a. Tnh th tch hnh chp cho.

2) Trong khng gian vi h to cc Oxyz vi h to vung gc Oxyz, cho

hai ng thng: 1:3

3

2

2

1

1 == zyx 2:

=+

=+

0532

02

zyx

zyx

Tnh khong cch gia hai ng thng cho.

Cu5: ( 1 im)

Chng minh rng: P1 + 2P2 + 3P3 + ... + nPn = Pn + 1 - 1

Trong n l s t nhin nguyn dng v Pn l s hon v ca n phn t.

s 48

Cu1: (3 im)

Cho hm s: y = x3 + 3x2 + 1 (1)


2) ng thng (d) i qua im A(-3 ; 1) c h gc l k. Xc nh k (d) ct th

hm s (1) ti ba im phn bit.Cu2: (2,5 im)

1) Gii phng trnh: 0221 =++++ xcosxsinxcosxsin

2) Gii h phng trnh:( )( )

=++

=++

095

1832

2

2

yxx

yxxx


50/155



Cu3: (2 im)


4 1+ xlogxlog 1

2) Tm gii hn:xcos

xxlimx

++ 1

121323 2

0

Cu4: (1,5 im)

Trong mt phng vi h ta cc Oxy cho hai im A(1; 2), B(3; 4). Tm trn

tia Ox mt im P sao cho AP + PB l nh nht.

Cu5: (1 im)

Tnh tch phn: I = +

+2

03 23

1dx

x

x

s 49

Cu1: (2,5 im)

Cho hm s: y = ( ) ( ) 4313

1 23 +++ xmxmx (1) (m l tham s)


2) Xc nh m hm s (1) ng bin trong khong: 0 < x < 3

Cu2: (2 im)

1) Gii phng trnh: 0322212 333 =+++++ xxx (1)

2) Cho phng trnh: ( ) 061232 2 =++ mxcosxsinmxsin

a) Gii phng trnh vi m = 1.


51/155



b) Vi gi tr no ca m th phng trnh (1) c nghim.

Cu3: (1 im)

Gii h bt phng trnh:

>+


52/155



Xc nh n du "=" xy ra?

Cu3: (2 im)

1) Cho phng trnh: xsinmxcosxsin 266 =+ a) Gii phng trnh khi m = 1.

b) Tm m phng trnh c nghim.

2) Chng minh rng ABC u khi v ch khi

++=

=

acbacb

a

Ccosba333

2

2

Cu4: (2,5 im)

1) Trong mt phng vi h ta cac Oxy cho im A(8; 6). Lp phng trnh

ng thng qua A v to vi hai trc to mt tam gic c din tch bng 12.

2) Trong khng gian vi h to cc Oxyz Cho A(1; 2; 2), B(-1; 2; -1),

C(1; 6; -1), D(-1; 6; 2)a) Chng minh rng ABCD l hnh t din v tnh khong cch gia hai ng

thng AB v CD.

b) Vit phng trnh mt cu ngoi tip t din ABCD.

Cu5: (1,5 im)

Cho hai hm s f(x), g(x) xc nh, lin tc v cng nhn gi tr trn on [0; 1].

Chng minh rng: ( ) ( ) ( ) ( )

1

0

1

0

21

0

dxxgdxxfdxxgxf

s 51

Cu1: (2 im)

Cho hm s: y =( )( )

mmxmxxm

+++ 421 2

(Cm) (m l tham s, m 0, -4

1)

1) Kho st s bin thin v v th ca hm s (C2) vi m = 2.

2) Tm m hm s (Cm) c cc i, cc tiu v gi tr cc i, cc tiu cng du.

Cu2: (2 im)

1) Gii h phng trnh:

++=

++=

22

22

3

3

yxy

xyx


53/155



2) Gii phng trnh: tg2x + cotgx = 8cos2x

Cu3: (2,5 im)

1) Tnh th tch ca hnh chp S.ABC bit y ABC l mt tam gic u cnh a,

mt bn (SAB) vung gc vi y, hai mt bn cn li cng to vi y gc .


(D1):

=+=+

0104

0238

zy

zx(D2):

=++=

022

032

zy

zx

a) Vit phng trnh cc mt phng (P) v (Q) song song vi nhau v ln lt i qua

(D1) v (D2).

b) Vit phng trnh ng thng (D) song song vi trc Oz v ct c hai ng

thng (D1), (D2)

Cu4: (2 im)

1) Tnh tng: S = ( ) nnn

nnnn nC....CCCC 14324321 +++

Vi n l s t nhin bt k ln hn 2, knC l s t hp chp k ca n phn t.


2

1 12xx

dx

Cu5: (1,5 im)

Cho ba s bt k x, y, z. Chng minh rng:

222222 zyzyzxzxyxyx +++++++

s 52

Cu1: (2 im)

Cho hm s: y =1

1

+

xx

(1) c th (C)

1) Kho st s bin thin v v th ca hm s (1).2) Chng minh rng ng thng d: y = 2x + m lun ct (C) ti hai im A, B

thuc hai nhnh khc nhau. Xc nh m on AB c di ngn nht.

Cu2: (2,5 im)

Cho phng trnh: 03232322

224 =+ . xx (1)


54/155




2) Xc nh m phng trnh (1) c nghim.

Cu3: (2,5 im)

Gii cc phng trnh v bt phng trnh sau:

1) xtgxsinxcos

xcosxsin 28

1322

66=

+

2) ( ) ( )2431243 2329 ++>+++ xxlogxxlog Cu4: (1,5 im)

Trong khng gian vi h to cc Oxyz Cho A(1; 1; 1), B(1; 2; 0) v mt

cu (S): x2 + y2 + z2 - 6x - 4y - 4z + 13 = 0. Vit phng trnh mt phng cha ng

thng AB v tip xc vi (S).Cu5: (1,5 im)

Tnh tng: S = nnnnn Cn

...CCC1

1

3

1

2

1 211

+++++

Bit rng n l s nguyn dng tho mn iu kin: 7921 =++ nnnn

nn CCC

knC l s t hp chp k ca n phn t.

s 53

Cu1: (2 im)

Cho hm s: y = -x3 + 3x2 - 2

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

2) Tm t phng trnh: 023 223 =+ tlogxx c 6 nghim phn bit.

Cu2: (3 im)

1) Trong mt phng vi h ta cc Oxy cho ng trn


55/155



(C): ( ) ( ) 413 22 =+ yx . Vit phng trnh tip tuyn ca (C) bit rng tip

tuyn ny i qua im M0(6; 3)

2) Trong khng gian vi h to cc Oxyz cho hnh hp ABCD.A'B'C'D'

Vi A(2; 0; 2), B(4; 2; 4), D(2; -2; 2) v C'(8; 10; -10).

a) Tm to cc nh cn li ca hnh hp ABCD.A'B'C'D'.

b) Tnh th tch ca hnh hp ni trn.

Cu3: (2 im)

1) Gii phng trnh: 21 +=++ xxx

2) Gii h phng trnh:

=

=+

22

1

22 yyx

x

ysinxsin

Cu4: (2 im)

1) Chng minh rng: knkn

kn

kn CCCCCCC =++

22

22

12

122

02

n k + 2 ; n v k l cc s nguyn dng, knC l s t hp chp k ca n phn t.

2) Tnh din tch hnh phng gii hn bi parabol: y = -x2 - 4x; ng thng x = -1;

ng thng x = -3 v trc Ox

Cu5: (1 im)

Cho 2 s nguyn dng m, n l s l

Tnh theo m, n tch phn: I =

2

0

xdxcosxsin mn

s 54Cu1: (2 im)

1) Kho st s bin thin v v th ca hm s: y = xxx 323

23 +

2) Da v th (C) Cu trn, hy bin lun theo tham s m s nghim ca

phng trnh: meee xx

x=+ 32

3

23

Cu2: (3 im)


56/155



1) Trong mt phng vi h ta cc Oxy cho elp (E) c phng trnh:

12

2

2

2

=+b

y

a

x(a > 0, b > 0)

a) Tm a, b bit Elip (E) c mt tiu im l F1(2; 0) v hnh ch nht c s ca (E)

c din tch l 12 5 (vdt).b) Tm phng trnh ng trn (C) c tm l gc to . Bit rng (C) ct (E) vatm c Cu trn ti 4 im lp thnh hnh vung.

2) Trong khng gian vi h to cc Oxyz tm theo a, b, c (a, b, c 0) to cc nh ca hnh hp ABCD.A'B'C'D'. Bit A(a; 0; 0); B(0; b; 0) C(0; 0; c) vD'(a; b; c).Cu3: (2 im)

1) Gii v bin lun phng trnh sau theo tham s m:

( ) 012333

= mlogxlogxlog

2) Gii phng trnh: ( ) 032332 =++++ xcosxcosxcosxsinxsinxsin Cu4: (2 im)

1) Cho f(x) l hm lin tc trn on [0; 1]. Chng minh rng:

( ) ( )

=2

0

2

0

dxxcosfdxxsinf

2) Tnh cc tch phn:

I =

+

2

020032003

2003

xcosxsin

xdxsinJ =

+

2

020032003

2003

xcosxsin

xdxcos

Cu5: (1 im)

Gii bt phng trnh: ( ) nnn

nnn C.C.C.!n 32

3 720knC l t hp chp k ca n phn t.

s 55

Cu1: (2 im)1) Kho st s bin thin v v th ca hm s: y = x4 - 10x2 + 9

2) Tm tt c cc gi tr ca tham s m phng trnh: x - 3mx + 2 = 0 c

nghim duy nht.

Cu2: (2 im)

1) Tm tt c cc ng tim cn xin ca th hm s: y = 2x + 21 x+


57/155



2) Tnh th tch ca vt th trn xoay c to ra khi cho hnh phng gii hn

bi cc ng: y = ex ; y =e1

; y = e v trc tung quay xung quanh Oy.

Cu3: (2 im)

1) Cho a thc: P(x) = ( )

2005

1516 x , khai trin a thc di dng:P(x) = 20052005

2210 xa...xaxaa ++++

Tnh tng: S = 2005210 a...aaa ++++

2) Gii h phng trnh:( )

=+=

5

115223

22 logyxlog

yx

Cu4: (2 im)

1) Cho ABC c di cc cnh BC, CA, AB theo th t lp thnh cp s

cng. Tnh gi tr ca biu thc: P =22CgcotAgcot

2) Trong mt phng vi h to cc vung gc Oxy cho hypebol (H):

1916

22

= yx . Lp phng trnh ca elp (E), bit rng (E) c cc tiu im l cc tiu

im ca (H) v (E) ngoi tip hnh ch nht c s ca (H)

Cu5: (2 im)

1) Trong khng gian vi h to cc Oxyz cho ABC c im B(2; 3; -4),

ng cao CH c phng trnh:52

2

5

1

== zyx v ng phn gic trong gc A l

AI c phng trnh:2

1

1

3

7

5 +== zyx . Lp phng trnh chnh tc ca cnh AC.

2) CMR: trong mi hnh nn ta lun c:2

6

V

3

3

2

S

(V l th tch hnh nn, S l din tch xung quanh ca hnh nn)

s 56Cu1: (2 im)

Cho hm s: y =( )

1

112

+++

xmxmx

(1) (m l tham s)



58/155



2) Chng minh rng hm s (1) lun c gi tr cc i (yC) v gi tr cc tiu

(yCT) vi m. Tm cc gi tr ca m (yC)2 = 2yCT

Cu2: (2 im)

1) Gii phng trnh: 3cosx ( ) 1221 2 = xsinxsinxcosxsin

2) Gii h bt phng trnh:

+

045

02

24

2

xx

xx

Cu3: (2 im)

1) Tnh tch phn: I = +3

0

231 dxxx

2) Tm s nguyn dng n tho mn ng thc: nCA nn 16223 =+

Cu4: (3 im)

1) Cho t din ABCD c di cnh AB = x (x > 0), tt c cc cnh cn li c

di bng 1. Tnh d di on vung gc chung ca hai cnh AB v CD. Tm iu

kin i vi x Cu ton c ngha.

2) Trong khng gian vi h to cc Oxyz cho t din OABC c O l gc

ta , A Ox, B Oy, C Oz v mt phng (ABC) c phng trnh:

6x + 3y + 2z - 6 = 0.a) Tnh th tch khi t din OABC.

b) Xc nh to tm v tnh bn knh ca mt cu ngoi tip khi t din OABC.

Cu5: (1 im)

Cho x, y l hai s thc dng khc 1.

Chng minh rng nu: ( ) ( )yloglogxloglog xyyx = th x = y.

s 57

Cu1: (2 im)

Cho hm s: y =2

52

xx


59/155



1) Kho st s bin thin v v th ca hm s.

2) Vit phng trnh tip tuyn ca th hm s, bit tip tuyn i qua im

A(-2; 0).

Cu2: (3 im)

1) Gii phng trnh: xsinxsin 243 = +

2) Gii bt phng trnh: ( ) ( )1111

2 +>+ xlogxlog xx

3) Gii h phng trnh:

=

=+

72

3432

22

22

yx

xyyx

Cu3: (2 im)

1) Tnh tch phn: ++

2

02

3

12 dxxx

x

2) Tm h s ln nht ca a thc trong khai trin nh thc Niutn ca:15

3

2

3

1

+ x

Cu4: (3 im)

1) Cho hnh lp phng ABCD.A'B'C'D'. Chng minh rng cc im gia ca 6

cnh khng xut pht t hai u ng cho AC' l nhng nh ca mt lc gic

phng u.

2) Trong mt phng vi h ta cc Oxy cho hai ng thng:x + y - 1 = 0 v 3x - y + 5 = 0

Hy tm din tch hnh bnh hnh c hai cnh nm trn hai ng thng cho, mt

nh l giao im ca hai ng v giao im ca hai ng cho l I(3; 3).


d1:

=+=+053

0523

zy

yxv d2:

25

2

1

2

=+= zyx

Chng minh rng hai ng thng cho nhau v tm phng trnh ng vunggc chung ca chng.

s 58

Cu1: (4 im)

Cho hm s: y =mx

mx

+ 13(1)


60/155



1) Xc nh m hm s (1) nghch bin trong khong (1; + )

2) Kho st s bin thin v v th ca hm s (1) khi m = 1, gi th ca

hm s ny l (C).

3) Tm hai im A, B thuc (C) sao cho A v B i xng vi nhau qua ng

thng (d): x + 3y - 4 = 0.Cu2: (2 im)

Cho phng trnh: x2 - 2ax + 2 - a = 0 (1)

1) Xc nh a phng trnh (1) c hai nghim x1, x2 sao cho: -2 < x1 < 3 < x2

2) Xc nh a phng trnh (1) c hai nghim x1, x1 sao cho:22

21 xx + t gi

tr nh nht.

Cu3: (1 im)

Cho ABC c 3 gc tho mn iu kin sau: sinA + cosA + sinB - cosB + sinC

- cosC = 1. Chng minh rng: ABC l tam gic vung.

Cu4: (3 im)

Cho ABC c A(-1; 5) v phng trnh ng thng BC: x - 2y - 5 = 0 (xB 0). Dng hnh hp ch nht nhn O, A, B, C

lm bn nh v gi D l nh i din vi nh O ca hnh hp .

1) Tnh khong cch t im C n mt phng (ABD).

2) Tnh to hnh chiu vung gc ca C xung mt phng (ABD). Tm iukin i vi a, b, c hnh chiu nm trn mt phng (xOy)

Cu5: (2 im)

1) Tnh tch phn: +

1

0 1xe

dx

2) Tnh h nguyn hm ca: f(x) = x(1 - x)20

s 76

Cu1: (2 im)

1) Kho st s bin thin v v th ca hm s: y = x3 - x2 - x + 1

2) Bin lun theo tham s m s nghim ca phng trnh: ( ) mxx =+ 112

Cu2: (2 im)

Gii cc phng trnh:

1) sin4x + cos2x + 4cos6x = 0

2) xlogx

logx

logxlogxlog xx 244244

22

222 =+++

Cu3: (1 im)

Tm tt c cc gi tr ca tham s m phng trnh sau c nghim:( )( ) mxxxx =+++ 2222

Cu4: (1,5 im)

Cho t din SABC vi gc tam din nh S l vung. Gi H l trc tm ca

ABC. Chng minh rng:

1) SH (ABC).


79/155



2)2222

1111

SCSBSASH++=

Cu5: (2 im)

Cho n N

1) Tnh tch phn: ( ) +1

0

21 dxxx n

2) Chng minh rng:1

12

1

1

4

1

3

1

2

11

1321

+=

++++++

+

nC

n...CCC

nnnnnn

Cu6: (1,5 im)

1) Tnh tch phn: I = ( ) +1

0

321 dxxx

n(n N)

2) Lp phng trnh ng thng i qua im M(1; 0) sao cho ng thng

cng vi hai ng thng: (d1): 2x - y + 1 = 0 (d2): x + 2y - 2 = 0 to ra mt tam gic

cn c nh l giao im ca hai ng thng d1, d2.

s 77

Cu1: (2 im)

Cho hm s: y = x3 + 3mx2 + 3(m2 - 1)x + m3 - 3m1) Kho st s bin thin v v th ca hm s ng vi m = 0.

2) Chng minh rng vi mi m hm s cho lun lun c cc i v cc tiu;

ng thi chng minh rng khi m thay i cc im cc i v cc tiu ca th

hm s lun lun chy trn hai ng thng c nh.

Cu2: (2 im)

1) Gii phng trnh lng gic:

sinx + sin2x + sin3x + sin4x = cosx + cos2x + cos3x + cos4x2) Chng minh rng trong ABC ta c:

+++=++

2222222

1111 Cgcot

Bgcot

Agcot

Ctg

Btg

Atg

CsinBsinAsin

Cu3: (2 im)


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1) Gii h phng trnh:

=+

=+

13

5

4224

22

yyxx

yx

2) Vi nhng gi tr no ca m th phng trnh: 15

1 2434

2

+=

+mm

xx

c bn nghim phn bit.

Cu4: (2 im)

Cho gc tam din ba mt vung Oxyz. Trn Ox, Oy, Oz ln lt ly cc im

A, B, C.

1) Tnh din tch ABC theo OA = a

2) Gi s A, B, C thay i nhng lun c: OA + OB + AB + BC + CA = k

khng i. Hy xc nh gi tr ln nht ca th tch t din OABC.Cu5: (2 im)

1) Tm h nguyn hm ca hm s: f(x) = tg4x

2) Tm h nguyn hm ca hm s: f(x) =xx

x

3

42

.

s 78Cu1: (2 im)

Cho hm s: y = f(x) = x4 + 2mx2 + m (m l tham s)1) Kho st s bin thin v v th ca hm s khi m = -1.

2) Tm tt c cc gi tr ca m hm s f(x) > 0 vi x. Vi nhng gi tr cam tm c trn, CMR hm s: F(x) = f(x) + f'(x) + f"(x) + f"'(x) + f(4)(x) > 0 xCu2: (2 im)

1) Gii phng trnh lng gic:( )

1

2

2

1

=

+ gxcotxsinxcos

xgcottgx

2) Hai gc A, B ca ABC tho mn iu kin: 122

=+ BtgAtg . Chng minh

rng: 124

3


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Trong khng gian vi h to cc Oxyz cho ng thng (d):

==+=

tz

ty

tx

3

2

21

v mt phng (P): 2x - y - 2z + 1 = 01) Tm to cc im thuc ng thng (d) sao cho khong cch t mi

im n mt phng (P) bng 12) Gi K l im i xng ca I(2; -1; 3) qua ng thng (d). Hy xc nh

to im K.Cu4: (2 im)


1265

3

1

3

12

3 +>++ xlogxlogxxlog

2) Vi a > 1 th phng trnh sau v nghim:

112222

++=++ aaxcosxxsinxCu5: (2,5 im)

1) Tnh din tch ca hnh phng gii hn bi parabol (P) c phng trnh:y = x2 - 4x + 5 v hai tip tuyn ca (P) k ti hai im A(1; 2) v B(4; 5)

2) Tnh tch phn: I = ( )

+2

0

442 dxxcosxsinxcos J =

0

dxxsinxcos

3) Vit khai trin Newton ca biu thc (3x - 1)16. T chng minh rng:

161616216141161501616 2333 =++ C...CCC s 79

Cu1: (2 im)Cho hm s: y = -x4 + 2(m + 1)x2 - 2m - 11) Xc nh tham s m th hm s ct trc honh ti bn im lp thnh

mt cp s cng.2) Gi (C) l th khi m = 0. Tm tt c cc im thuc trc tung sao cho t

c th k c ba tip tuyn vi th (C).Cu2: (2 im)

1) Gii phng trnh: x2 + 11 =+x

2) Gii v bin lun phng trnh: m.cotg2x =xsinxcos

xsinxcos66

22

+

theo tham s m

Cu3: (1,5 im)1) Cho hai hm s: f(x) = 4cosx + 3sinx; g(x) = cosx + 2sinx

a) Tm cc s A, B tho mn: g(x) = A.f(x) + B.f'(x)


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b) Tnh tch phn:( )

( )

4

0

dxxfxg

2) Tm th tch vt th to bi elp:( )

1164

422

+ yx quay quanh trc Oy

Cu4: (2,5 im)1) Cho hnh hp ch nht ABCD.A1B1C1D1; H v K l cc hnh chiu vung

gc ca A v C1 xung mt phng (B1CD1). Chng minh: 12KCAH = 2) Cho hai ng trn: tm A(1; 0) bn knh rA = 4 v tm B(-1; 0) bn knh rB

= 2. Tm tp hp tm I(x, y) ca cc ng trn tip xc c 2 ng trn trn. Tp hp l ng g?

3) Vit phng trnh ng thng d vung gc vi mt phng (P): x + y + z = 1

v ct c hai ng thng d1:

11

1

2

1 zyx =

+= d2:

=++

=+

0122

042

zyx

xyx

Cu5: (2 im)1) Cho ba hp ging nhau, mi hp ng 7 bt ch khc nhau v mu sc.

Hp I c 3 bt mu , 2 bt mu xanh, 2 bt mu en;Hp II c 2 bt mu , 2 bt mu xanh, 3 bt mu en;Hp III c 5 bt mu , 1 bt mu xanh, 1 bt mu en;

Ly ngu nhin mt hp v rt h ho t hp ra 2 bt.a) Tnh tt c s cc kh nng xy ra v s kh nng 2 bt c cng mub) Tnh s kh nng 2 bt khng c mu en

2) C bao nhiu s t nhin khc nhau, nh hn 10.000 c to thnh t 5 chs: 0, 1, 2, 3, 4

s 80

Cu1: (2,5 im)

Trong mt phng vi h ta cc Oxy cho (C) l th ca hm s

y = x +x1

v (d) l ng thng c phng trnh y = ax + b

1) Tm iu kin ca a v b (d) tip xc vi (C).

2) Gi s (d) tip xc vi (C) ti I. Gi M v N theo th t l giao im ca (d)

vi trc tung v vi ng phn gic ca gc phn t th nht. Chng minh:

a) I l trung im ca on MN.

b) Tam gic OMN c din tch khng ph thuc vo a v b.


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Cu2: (1,5 im)

Tm k h phng trnh:

==+

kyx

yx 122 c nghim duy nht.

Cu3: (1,5 im)

1) Chng minh rng: 11 22 ++++ aaaa 2 a R

2) Gii h phng trnh:

=+

=

1023

122

xyyx

xyyx

Cu4: (3 im)

1) Tm h nguyn hm ca hm s: f(x) = (sin4x + cos4x)(sin6x + cos6x)

2) Trong mt phng vi h ta cc Oxy cho hai ng thng:

(1): 4x - 3y - 12 = 0 (2): 4x + 3y - 12 = 0a) Tm to cc nh ca tam gic c ba cnh ln lt nm trn cc

ng thng (1), (2) v trc tung.

b) Xc nh tm v bn knh ng tr

BỘ ĐỀ THI THỬ ĐẠI HỌC 2009 CHỌN LỌC (toanhoccapba.wordpress.com)

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