Cac Chuyen de Luyen Thi Dai Hoc

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Chuyn :Phng trnh v bt phng trnh i sMt s dng h ph-ng trnh th-ng gp1) H ph-ng trnh bc nht: Cch tnh nh thc 2) H ph-ng trnh i xng loi 1: H khng thay i khi ta thay x bi y v ng-c li 3) H ph-ng trnh i xng loi 2: Nu i vai tr ca x v y th ph-ng trnh ny tr thnh ph-ng trnh kia v ng-c li 4) H ph-ng trnh ng cp bc 2: Xt 2 tr-ng hp, sau t x = ty 5) Mt s h ph-ng trnh khc

Cc v dV d 1. Mt s h dng c bn xy( x 1)( y 1) 1) Cho h ph-ng trnh x y x2 y2 a) Gii h khi m = 12 b) Tm m h c nghim 1 1 a 2) Cho h ph-ng trnh x ym 8

x2 y 2 a2 2 Tm a h ph-ng trnh c ng 2 nghim phn bit x 2 xy y 2 1 3) Cho h ph-ng trnh Tm m h c nghim x 2 3 xy 2 y 2 m

4) Cho h ph-ng trnh

x x2

y y2

a 6 a2

a) Gii h khi a = 2 b) Tm GTNN ca F = xy + 2(x + y) bit (x, y) l nghim ca h ( y 1) 2 m x 5) Cho h ph-ng trnh Tm m h c nghim duy nht ( x 1) 2 m y 6) Gii h ph-ng trnh:

x y

2

y

2 2 3 x 1 y 1 m

2 x y 1

7) Gii h ph-ng trnh:

x 1 x y 1

y x 1

a) Gii h khi m = 6 b) Tm m h c nghim3y y2 2 x2 (KB 2003) x2 2 y2

V d 2. Gii h ph-ng trnh:3x

HD:

TH1 x = y suy ra x = y = 1 TH2 ch : x>0, y> 0 suy ra v nghim2x 2 y 8x3

V d 3. Gii h ph-ng trnh:

xy 2 y3

15 35

Convert by TVDT

1

www.luyenthi24h.com www.luyenthi24h.comHD: Nhm nhn t chung sau tx3V d 4. Gii h ph-ng trnh:

S = 2x + y v P = 2x. yy3 1 3y (2) (1)

s: (1, 3) v (3/2, 2)

3x y6

x6

HD: t (2) : - 1 x, y 1 hm s: f t

t3

3t trn [-1;1] p dng vo ph-ng trnh (1)2x2

y x

V d 5. CMR h ph-ng trnh sau c nghim duy nht:2y2

a2 y a2 x

HD:

x 2x3

y x2

a

2

;

xt f ( x)

2x 32

x 2 , lp BBT suy ra KQy 2

V d 6. Gii h ph-ng trnh:

x

y 2 x 2 HD Bnh ph-ng 2 v, i xng loi 2 xy x 2 a ( y 1) V d 7. xc nh a h c nghim duy nht xy y 2 a ( x 1)

HD s dng K cn v V d 8. Gii h ph-ng trnh:

a=8xy 10 xy 5 20 x 2 (1) y 2 (2)5 y y 2 5 ; x2

HD: Rt ra x3

5 y

y2

5 y

y ; C si x

20 theo (1)

x2

20 suy ra x, y

V d 9.x

x y

y x

x

y y 2

(1)

(KB 2002) HD: t (1) t cn nh lm nhn t chung

(1;1) (3/2;1/2)

V d 10.

x 1 x y 3a

y

2

a

Tm a h c nghim

HD: T (1) t u x 1, v y 2 -c h di xng vi u, -v Ch ra h c nghim th ph-ng trnh bc hai t-ng ng c 2 nghim tri du

Bi tp p dng 1) 2) 3)4)6x 2 5x 2 xy 2 y 2 xy y2 56 49

x2 x2(x2 x2 x3 x2

x y2

y2 3( x

y y)

KD 2003

2 x)(3 x 5x y3 y2 y 9 7( x x

y ) 18 0 y) y 2

HD: tch thnh nhn tConvert by TVDT

4 nghim2

www.luyenthi24h.com www.luyenthi24h.comxy x2

5) 6) 7)

y2 xy

12 26 m 2

Tm m h c nghim

(x xx23

y) 2 . y y3

19y) 6 9

t t = x/y

H pt c 2 nghim

x( x 2)( 2 x 4x y

t X = x(x + 2) v Y = 2x + y(1) 4

8)9)

x x2

y y2

x

y x2

2 y2

HD: i bin theo v, u t ph-ng trnh (1)

1 x3 y3 y xy 2

19x 3 6x 2

HD: t x = 1/z thay vo -c h y, z S ( - 1/2, 3) (1/3, - 2)

10)

x 2y

1 x

y x3 1

1 y (KA 2003)

HD: x = y V xy = - 1 CM x 4 x 2 0 v nghim bng cch tch hm s 11)( x 1) 2 ( y 1)2

kq: 3 nghim HD s dng K cn v

y

a

x a

xc nh a h c nghim duy nht

12)x

2x y y

2y x xy 3

3

HD bnh ph-ng 2 v

13)

x y x xy

y x y xy

7 xy 78

1

HD nhn 2 v ca (1) vi

xy

2. Ph-ng trnh v bt ph-ng trnh ph-ng trnh i sMt s dng ph-ng trnh v bt ph-ng trnh th-ng gp1) Bt ph-ng trnh bc hai nh l v du ca tam thc bc hai Ph-ng php hm s 2) Ph-ng trnh, bt ph-ng trnh cha gi tr tuyt i A B A2 B 2

A

B

A A

B B B

0 0)

A B B A 3) Ph-ng trnh, bt ph-ng trnh cha cn thc

B (B

Mt s v dV d 1. Tm m ( x 1)(x 3)(x 2

4 x 6)

m nghim ng vi mi xConvert by TVDT 3

www.luyenthi24h.com www.luyenthi24h.comHD: s dng hm s hoc tam thc: m - 2 V d 2. Tm a h sau c nghimx y y x 2 2 x( y 1) a(1) ( y 2)2

2

HD:

x

y 22

( x 1)

a 1

(2)

TH1: a + 1 0 H v nghim TH2: a + 1>0. V th (2) l -ng trn cn (1) l min gch cho: a - 1/2V d 3. Gii cc ph-ng trnh, bt ph-ng trnh sau 1) 2)

8x 2 6 x 1 4 x 1 0 x 4 1 x 1 2x : x = 02 x) x2 2x 3 9 0 x 1 5

3) 2( x 2 4)

x

x2 13x) x 2

x3x 2

x2 1

2

HD: Tch 2 nhn t bng 1 suy ra cch giix 2 10x 9 x2

5) ( x 2

0 KD 20020 0 2x 1 m

V d 4. Tm m h sau c nghim

S: m4

V d 5. Gii bt ph-ng trnh 2 x 1 2 x x 2 HD + / Nhn 2 v vi biu thc lin hp ca VT

+ / Bin i v BPT tch ch KV d 6. Gii bt ph-ng trnh: 3 x HD t t

3 2 x

2x

1 2x

7

x

1 2 x

,t

2 , AD BT c si suy ra Kx2 (1 x 1) 2 x 4

V d 7. Gii bt ph-ng trnh:

HD: + / Xt 2 tr-ng hp ch y DK x> = - 1 + / Trong tr-ng hp x 4, tin hnh nhn v chia cho biu thc lin hp mu VT V d 8. Cho ph-ng trnh: x 9 x HD: + / Bnh ph-ng 2 v ch K

x2

9 x m . Tm m ph-ng trnh c nghim

+ / t t = tch 2 cn thc, Tm K ca t + / S dng BBT suy ra KQV d 9. Gii bt ph-ng trnh (KA 2004) :2( x 2 16) x 3 x 3 7 x x 3

Bi tp p dng1)x2 x y y2 a 2x 1 0

Tm a h c nghim duy nht. Tm nghim duy nht .

S a = - 1 v a = 3 2) Tm m bt ph-ng trnh sau c nghim: 3) 4)

4x 2

16 4 x

m

x 4x 12

x 4x 3

2 x 12 2 x 2 162x 1Convert by TVDT 4

www.luyenthi24h.com www.luyenthi24h.com5) 2(1 x) x 26) 7)( x 1) x

2x 1(2 x) x

x2

2x 1

HD: t t

x2

2x 1 , coi l ph-ng trnh bc hai n t

2 x2

x 2 x 1

( x 2) x 1

x 3 2x x 4 m

8) Cho ph-ng trnh:

x 4 x 4

a) Gii ph-ng trnh khi m = 6 b) Tm m ph-ng trnh c nghim 9)51 2 x x 2 1 x3x 4

12x 3 2 0

10) x 2

11) Tm a vi mi x: f ( x)

( x 2) 2

2x a

3 S a 4 ; a 0

Chuyn 3:Mt s kin thc cn nh

L-ng gic

1. Ph-ng trnh v h ph-ng trnh l-ng gicCc cng thc bin i l-ng gic Mt s dng ph-ng trnh c bn Ph-ng trnh bc 2, bc 3 theo mt hm s l-ng gic Ph-ng trnh ng cp bc nht vi sinx, cosx: asinx + bcosx = c Ph-ng trnh ng cp bc 2 vi sinx, cosx: a. sin2x + b. sinx. cosx + c. cos2x + d = 0 Ph-ng trnh ng cp bc 3 vi sinx, cosx: a. sin3x + b. sin2x. cosx + c. sinx. cos2x + d. cos3x = 0 a. sin3x + b. sin2x. cosx + c. sinx. cos2x + d. cos3x + m = 0 Ph-ng trnh i xng vi sinx, cosx a: (sinxcosx) + b. sinx. cosx + c = 0 Ph-ng trnh i xng vi tgx, cotgx Ph-ng trnh i xng vi sin2nx, cos2nx

Cc v dV d 1. V d 2.

cot xcos2 x

tan x

2.cos 4 x sin 2 xcos2 x 2 3

HD: t K x = /3 + k.1 (sin x 1) 2

3

HD: S dng cng thc h bc 1 2. cos(2 xV d 3. V d 4.

).cos

3

sin x

S 3 h nghim

sin 2 x sin 2 2 x 2 sin 2 2 x sin 2 x sin 3 x.sin 3x cos3 x.cos 3xtan x 6 .tan x 3

HD: Nhm, nhn ln v tch 2 thnh 2 nhm1 8

HD: t K rt gn MS = 1;V d 5. 3 tan x(tan x 2.sin x) 6.cos x

AD cng thc nhn 3; S x = - /6 + k0

HD: Bin i theo sin v cos -c 3. cos2 x(1 2 cos x) sin 2 x(1 2 cos x) 0Convert by TVDT

S x = /3 + k5

www.luyenthi24h.com www.luyenthi24h.com3.tan y 2 6sin x 2sin x 2sin( y 6sin( y x) x)

V d 6.tan

y 2

HD: nhn (1) vi (2) rt gn tan 2

y 2

4sin 2 y 1 . sin 3x 2

t t

tan 2

y ; 2

t = 0, t

3

V d 7. cos3x. sin 2 x cos 4 x. sin x

1 cos x HD: B tch thnh tng rt gn

1 2 HD: nhn 2 v vi 2. sin(x/2) ch y xet tr-ng hp bng 0 T cos x cos 2 x .. cos nx NX: Trong bi ton cha tng thc hin rt gn bng cch trn T sin x sin 2 x .. sin nxV d 8. cos x cos 2 x cos3x cos 4 x cos5 x V d 9. tan x.sin 2 x 2.sin 2 x V d 10. logcos 9 2 x

3(cos 2 x sin x.cos x)

HD: B sau t t = tg(x/2)logsin x 2 logsin x (sin x) 2 4

.4.log sin 2 x 2 4

HD: 2 logsin x .2.

2. Gi tr ln nht nh nht, ph-ng trnh c tham sMt s kin thc cn nh Ph-ng php hm s: Bi ton Max, Min trn 1 khong v mt on. Ph-ng php bt ng thc, nhn xt nh gi.

Cc v d3. cos4 x 4 sin 2 x 3. sin 4 x 2 cos2 x HD: t = cos2x, tm Max, Min trn 1 on M = 8/5 m = 4/3 V d 2. Cho ph-ng trnh: cos2 x m. cos2 x 1 tgx 1) Gii ph-ng trnh khi m = 1 2) Tm m ph-ng trnh c nghin thuc on [0; /3]V d 1. Tm GTLN, GTNN: y

HD: t = tgx, tV d 3. :

0; 3 ; Lp BBT f(t)

S: m

(1

3) 1

3 ;1

Tm GTLN, GTNN: y 2. sin8 x cos4 2 x HD: t = cos2x, - 1t1 tm Max, Min trn 1 on f , t 0 8t 3 (t 1) 3 S:M = 3, m = 1/27 V d 4. Tm GTLN, GTNN: y cos4 x sin 4 x sin x. cos x 1 V d 5. Cho ph-ng trnh: 2.(sin4 x cos4 x) cos 4 x 2 sin 2 x m 0 Tm m ph-ng trnh c t nht mt nghin thuc on [0; /2] S: [ -10/3; -2] 2 sin x cos x 1 V d 6. Cho ph-ng trnh a sin x 2 cos x 3 1) Gii ph-ng trnh khi a = 1/3 2) Tm a ph-ng trnh c nghim HD: -a v dng: (2 - a) sinx + (2a + 1) cosx = 3a + 1 S [ -1/2, 2] x 3 V d 7. Tm nghim ca pt sau trong khong (0, ) : 4 sin 2 3. cos 2 x 1 2 cos2 x 2 4

Bi tp p dngConvert by TVDT 6

www.luyenthi24h.com www.luyenthi24h.com1) cos x. cos 2 x. cos3x sin x. sin 2 x. sin 3x 2) sin x 3) 4)sin x 5 3sin 2 (3 x) 2sin x 2 1 2. sin 3x 2. cos3x sin x 1 cos 2 x 1 cot 2 x sin 2 2 x cos 2 x cos x(2.tan 2 x 1) 3. cos x 3. cos x.cos 2

1 23 2

2x 5sin 2 x 0

1 cos xHD: Ch K

5)

S: x = - /4 + k /2

6) 2 6 2 7) 3 cos 4 x 8 cos x 2 cos 3 0 x (2 3 ) cos x 2 sin 2 sin 3x 2 4 8) 2 cos x 1 9) 1 sin x cos x sin 2 x cos 2 x 0

1

Mt s thi t nm 20021) Tm nghim thuc khong 0;2 2) Gii ph-ng trnh 1 tan 4 x ca ph-ng trnh 5 sin xcos3x sin 3x 1 2 sin 2 x cos 2 x 3 KA 2002

(2 sin 2 2 x)sin 3x (DB 2002) cos4 xca ph-ng trnh cot 2 x tan x 4sin 2 x

2 KB 2003 sin 2 x 4) Tm x nghim ng thuc khong 0;14 ca ph-ng trnh cos 3x 4 cos 2 x 3cos x 4 0 KB 20033) Tm nghim thuc khong 0;2 5) Xc nh m ph-ng trnh 2 sin 4 x cos 4 xcos 4 x 2sin 2 x m 0 c t nht mt nghim thuc on

0;

2

(DB 2002)

1 1 (DB 2002) cot 2 x 2 8sin 2 x x 7) Gii ph-ng trnh tan x cos x cos 2 x sin x 1 tan x.tan (DB 2002) 2 2sin x cos x 1 8) Cho ph-ng trnh a (1) sin x 2cos x 3 1 a) Gii ph-ng trnh (2) khi a 3 b) Tm a ph-ng trnh c nghim 1 sin x (DB 2002) 9) Gii ph-ng trnh 8cos 2 x cos 2 x 1 10) Gii ph-ng trnh cot x 1 sin 2 x sin 2 x (KA 2003) 1 tan x 2 11) Gii ph-ng trnh 3 tan x tan x 2sin x 6cos x 0 (DBKA 2003)6) Gii ph-ng trnh 12) Gii ph-ng trnh cos 2 xcos x 2 tan 2 x 1 2 (DBKA 2003)

sin 4 x cos4 x 5sin 2 x

Convert by TVDT

7

www.luyenthi24h.com www.luyenthi24h.com13) Gii ph-ng trnh 3cos 4 x 8cos6 x 2cos 2 x 3 0 (DBKB 2003) x 2 3 cos x 2sin 2 2 4 1 (DBKB 2003) 14) Gii ph-ng trnh 2 cos x 1 x x 15) Gii ph-ng trnh sin 2 .tan 2 x cos 2 0 (KD 2003) 2 4 2 cos2 x cos x 1 16) Gii ph-ng trnh 2 1 sin x (DBKD 2003) cos x sin x 2sin 4 x 17) Gii ph-ng trnh cot x tan x (DBKD 2003) sin 2 x 18) Gii ph-ng trnh 5sin x 2 3 1 sin x tan 2 x (KB 2004) 19) Gii ph-ng trnh 2cos x 1 2sin x cos x

sin 2 x sin x (KB 2004)

Chuyn 4:Mt s kin thc cn nh Cc cng thc v m v lgarit. Gii thiu mt s ph-ng trnh c bn. Khi gii ph-ng trnh v logarit ch K.

M & Lgarit

1. Ph-ng trnh v h ph-ng trnh M lgarit

Cc v d2 2 V d 1. Cho ph-ng trnh: log3 x log3 x 1 2m 1 0 1) Gii ph-ng trnh khi m = 2 2) Tm m ph-ng trnh c t nht mt nghim thuc 1;3 3

HD: m [0;2]

V d 2.

s (4, 4) 2 log 4 x log 2 y 4 1 1 V d 3. log 2 ( x 3) HD: K x>0 V x1; S x = 2, x log4 ( x 1) 8 log2 (4 x) 2 4 V d 4. log5 x. log3 x log5 x. log3 x HD: i c s S: x = 1 v x = 159 log2 ( xy ) x2 y2 3( xy) log2 3 3y1)

log 2 ( x

2

y )

2

5

2 3 3

V d 5. V d 6.

3x

6

2 log3 ( x

x2 Suy ra 3y

HD: K x> - 1

TH1: - 10, t y = log3(x + 1)1 3y

1

V d 7. log 2

x2 1 x

3x 2

2x3

HD: VP 1 vi x>0, BBT VT 1 ; Csi trong lgagrit

S x = 1

23x

5y2

4yS (0, 1) (2, 4)

V d 8. 4 x 2 x 1 2x 2

y

Convert by TVDT

8

www.luyenthi24h.com www.luyenthi24h.comV d 9. Tm m ph-ng trnh sau c nghim thuc [32, + ) :

log2 x log 1 x 2 22

3

m log4 x 2

3

m

0, m 1 tlog x y 3

HD: t > = 5; 1 3m 2 m2 1 V d 10.

1 m

3

log y 2x

xy 2y

HD K x, y>0 v khc 1; B (1) -c TH1: y = x thay vo (2) c nghim 1 TH2: x thay vo (2) CM v nghim chia thnh 2 min y>1 v 03/2

a

0

6)

12 x 1) log 2 ( x 2 2 x)

10) Gii ph-ng trnh log 3 ( x 2 11)y 2 yx 1 2

x xx4 4

y

12)

(x 4 8( x 4

y ).3 y

1y

y) 6 x

02

13) Tm m ph-ng trnh 4 log2 x

log 1 x m2

0 c nghim thuc khong (0;1)

Chuyn 5. Tch phn xc nh v ng dng1. Ph-ng php tnh tch phnI. Tch phn cc hm s3

hu t0

V d : Tnh cc tch phn sau1) A 2) A1 1

x 2 .dx ; x) 9 2 (12

B1 4

x

2

dx ; 3x 2

2

5) A1 2

x

3

dx 2x 2

2

x

; B1

x

4

dx 4x 21

3

;

(x2

2 x 2.dx ; x3 13 2

B

x 3 dx ; 10 2 ( x 1)

6) A1

(x3

x 2 4 x 1).dx ;B x4 x3

x 3 .dx ; 8 4) 2 0 (x

A3)1 1

(2 x

10x 16x 1).dx ; x 2 5x 6

2

7) A13

dx x( x 6 1) 24

; B

(1 x 4 ).dx ; x.( x 4 1) 11 0

3

B4) A

dx ; 3) 2 ( x 1) 2 0 (x1

8) A0

x 5 dx ; B x6 x3 2 3 3

2 x 2 2 x 13 dx; ( x 2)( x 2 1) 2

(x3

1

3x 2 x 6).dx ;B x 3 5x 2 6 x

(7 x 4)dx ; 3 3x 2 1x

Convert by TVDT

10


Bi tp1) (CSP HN 2000): I

3x 2 2 .dx x2 0 11

3

2) (HNL TPHCM 1995) I0 1

x

2

dx 5x 6

3) (HKT TPHCM 1994) I

x .dx 3 0 (1 2 x )

4) (HNT HN 2000) 1 ( x 3 2 x 2 10 x 1).dx I x 2 2x 9 01

x 5 .dx 9) (HTM 1995) I 2 1 0 x 10) (H Thi Nguyn 1997) 2 (1 x 2 ).dx 1 I HD : t x 4 x x 1 1 11) Xc nh cc hng s A,B x 2 A B Tnh 2 2 x 1 ( x 1) ( x 1)3

1

I

( x 2) .dx 2 2 ( x 1)2

5) (HSP TPHCM 2000) I1

(4 x 11).dx 2 5x 6 0 x

12) Cho hm s f ( x)

6) (HXD HN 2000) I0 1

3.dx x3 14

dx 4x 2 3 0 x 8) (HQG HN 1995). Xc nh cc hng s A,B,C 3x 2 3x 3 A B C Tnh 3 2 x 1 x 2 x 3x 2 ( x 1)

7) (H MC 1995 ) I

x ( x 1) ( x 1) 3 a) nh cc h s A,B,C,D,E sao cho Ax 2 Bx C dx dx f ( x)dx D E 2 x 1 x 1 ( x 1)(x 2)3

b)

Tnh2

f ( x)dx

3x 2 3x 3 .dx x 3 3x 2 II Tch phn cc hm s l-ng gic V d : Tnh cc tch phn sau I1) Adx ; B 1 sin x cos x 02 3

tan x.dx 2 cos x sin x.cos x

3) A

( x sin x)dx ; B 1 cos x 02

4

2

sin 2 x. cos2 2 x.dx0

6

4) A3

2) A0

tan 4 x.dx ; B cos 2 x

3

( cos x6

sin x ).dx

0

x. cos x.dx ; 1 sin 2 x

Bi tp1) (HQG TPHCM 1998) Tnh :2 sin 2 x.dx sin 2 x.dx I ; va J 4 4 0 1 sin x 0 cos x 1 2) (HSP TPHCM 1995) sin x Cho f ( x) sin x cos x 2

a) Tm A,B sao cho cos x sin x f ( x) A B cos x sin x3

b) Tnh I0

f ( x).dx

3) (HGTVT TPHCM 1999)

Convert by TVDT

11


a) CMR

cos4 x.dx 4 sin 4 x 0 cos x2 2

sin 4 x.dx 4 sin 4 x 0 cos x2

2

10) (HQG TPHCM 1998) I0 4

cos3 x. sin 2 x.dx

cos4 x.dx 4 sin 4 x 0 cos x 4) (HTS 1999) Tnh :b) Tnh I2

I0

sin x. cos x.(1 cos x) 2 .dx dx 4 0 cos x4

4

sin 4 x.dx 2 0 1 cos x 12) (HBK HN 1999) Cho hm s sin 2 x h( x ) (2 sin x) 2 A. cos x B. cos x a) Tm A,B h( x) 2 2 sin x (2 sin x)11) (HVNH TPHCM 2000) I0

5) (HTM HN 1995) Tnh I

b) Tnh I2

h( x).dx

6) (HVKTQS 1999):Tnh I0

4. sin 3 x.dx 1 cos4 x

13) (HBK HN 1998)2

7) (HNN1 HN Khi B 1998) I2

cos 2 x.dx 1 cos x 0 sin 3 x.dx 2 0 1 cos x

2

I0

cos 2 x.(cos4 x sin 4 x).dx

14) (HVNH TPHCM 2000) I

8) (HQGHN Khi A 1997) I 9) (HNN1 HN 1998) Tnh2

( x sin x).dx cos2 x 0

3

I6

1 sin 2 x cos 2 x. .dx sin x cos x

III. 1) A

Tch phn cc hm s v t1 2a

V d : Tnh cc tch phn sau :x 15 . 1 3 x 8 .dx; B0 0

x. 2a

x 2 .dx (a

0)

1

6) A03

x dx4

7 2

2)a 4

x

3

; B0

dx3

1

2x 13

A0

x . a

2

2

x .dx; B1

2

dx x(1 x)

(a

0)

7) A8

dx x 1 x0 3 1

; (*)B0

( x 1 2)dx x2

2x

1 x 1

0

3) A1

dx x2

2

; B1

dx ( x 1)( x 2)

8) (*)A1

x 1 dx ; x 1x 10 2

x 12

1

4) A1 22

1 x .dx ; B x2dx x. x 2 1 ; B0

0

9) A0

4 x dx; B1

x21

2 x 2 .dx

dx x 42

12 2

x 21.dx

2

10) A1

x2 1 dx; B x

1 2

1 x2 .dx x2

5) A1

x x

Bi tp12

Convert by TVDT

www.luyenthi24h.com www.luyenthi24h.com1

1) (HVNH THCM 2000) I0

x 3 .dx x x2

2

6) (HSP2 HN 2000) I11 1

dx x. x 3 1

2

2) (H BKHN 1995) I2 31

dx x. x 2 1

7) (HXD HN 1996) I0

( x 2 1).dx x 1

7

3) (HVKTQS 1998) I1

dx 1 x x2

8) (HTM 1997) I10

x 3 .dx3

1 x21

4

4) (HAN 1999) I

dx 91

9) (HQG TPHCM 1998) I0

x.dx 2x 1

2 7 x. x

5) (HQG HN 1998) I0

x 3 . 1 x 2 .dx

IV. Mt s dng tch phn c bit V d1 :Tnh cc tch phn sau :6 sin xdx cos xdx B sin x cos x sin x cos x 0 0 V d2 :Tnh cc tch phn sau 4 1

1) A

2) A

e x .dx x e x 0 e1 2

4

B0

cos2 x. cos 2 x.dx

1

1) A

x 5 cos 2 x.dx; B1

x 3 e x .dx

2

2) A1 2

x 2 . ln

1 x .dx; B 1 x

2

sin 3 x 1 cos x

.dx

2

V d 3 :Tnh cc tch phn sau1) A

sin 2 x .dx; B 4 0 1 sin x

2

cos2004 x .dx 2004 x sin 2004 x 0 cos2

2) A

x. sin x .dx; B 3 cos 2 x 0

x. sin x .dx 1 cos 2 x 0

Bi tp1

1) (HPCCC 2000) Tnh I1

1 x2 .dx 1 2x

4) (HNT TPHCM 1994)Tnh I1

sin 2 x .dx 3x 1

2

2) (HGT 2000 )Tnh I2

x cos x .dx 4 sin 2 xx. sin 3 x.dx0

5) (HVBCVTHN 1999)Tnh I

x4 .dx 1 2x 1

3) (HQG HN 1994) Tnh I

2. ng dng ca tch phn xc nhMt s kin thc cn nhNi dung cc bi ton v din tch hnh phng: 3 bi ton c bn. Bi ton v th tch trn xoay.

Cc v dBi 1. Tnh th tch vt th trn xoay sinh ra bi php quay xung quanh trc ox ca hnh phng gii hn bi 2 sin x(0 x ). trc ox v -ng y Bi 2. Tnh din tch hnh phng gii hn bi cc -ng: yx2 4x 3, y x 3.

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13

www.luyenthi24h.com www.luyenthi24h.comx2 x2 . ,y 4 4 2 Bi 4. Tnh din tch hnh phng gii hn bi (P) y2 = 16x v cc tip tuyn ti A(1;4) B(4; - 8).Bi 3. Tnh din tc hnh phng gii hn bi cc -ng: y

4

Bi 1 Din tch phng1) (HBKHN 2000): Tnh din tch gii hn bi y 2) (HTCKT 2000): Tnh din tch gii hn bi y

sin 2 x. cos3 x; y

0 va x

0; x

20; x

e va x 1 3x 12x 3) (HVBCVT 2000) Tnh din tch gii hn bi y 1 2 sin 2 ;y 1 va x 2 4) (HVBCVT 1997) Tnh din tch gii hn bi y x 2 2 x; y 3x 2 2 5) (HTM 1996) Tnh din tch gii hn bi y x ; x y6) (HKT 1994) Tnh din tch gii hn bi y 7) (HC 1999) Tnh din tch gii hn bi yx2 4x 3; y 3 x

e ;y

x

x

2

x2; yx2

8) (HSP1 HN 2000) Tnh din tch gii hn bi y

x2 8 va y 8 x 1; y x 5

9) (HKTQD 1996) Tnh din tch gii hn bi hnh pha d-i (P) : y=ax2 (a>0) v trn y=ax+2a 10) Tnh din tch gii hn bi ( P) : y x 2 4 x 3 v 2 tip tuyn ti cc im A(0;-3) v B(3;0) 11) (H Hu 1999) Tnh din tch gii hn bi y ( x 1) 5 x; y e x va x 1 12) Tnh din tch gii hn bi y

sin 3 x; y

cos3 x va truc Oy voi 0

x

13) (HVQY 1997) Tnh din tch gii hn bi y cong (C) ti im c honh x=2 14) (HKT 2000) Tnh din tch gii hn bi y x=-1

0;4x x4

(C) : y

x

3

2x

2

4 4 x 3 v tip tuyn vi -ng

1

(C ) v Ox, hai -ng thng c ph-ng trnh x=1;

*****Mt s bi tham kho************ 1) Tnh din tch S gii hn bi th (C ) : y2) Tnh din tch S gii hn bi th (C ) : y

x 2 trc Ox v -ng thng c ph-ng trnh x=2 1 2 .x 2 trc Ox v 2 -ng thng c ph-ng trnh x=1 v 2

x=3 3) Tnh din tch S gii hn bi th (C ) : y x 2 trc Ox v -ng thng c ph-ng trnh x=2, y=x 4) Tnh din tch S gii hn bi th ( P) : y 2 2 x v -ng thng c ph-ng trnh y=2x-2 5) Tnh din tch S gii hn bi th ( P ) : x 2 y 2 va (P2 ) : x 1 3 y 2 1

Bi 2 Th tch ca cc vt th 1) (HNN1 HN 1997): Cho hnh phng gii hn bi Dy tgx; x 0; x 3 ;y 0

a) Tnh din tch hnh phng gii hn bi D b) Tnh th tch vt th trn xoay khi D quay quanh Ox 2) Tnh th tch ca vt th trn xoay sinh ra bi php quay quanh Ox ca hnh gii hn bi trc Ox v (P) y=x2-ax (a>0) 3) (HXD 1997) Tnh th tch ca vt th trn xoaydo hnh phng S y x. ln x; y 0; x 1; x e

4) (HY 1999) Tnh th tch hnh trn xoay sinh ra bi ( E ) :

x2 a2

y2 b2

1 khi n quay quanh Ox14

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5) (HTS TPHCM 2000): Cho hnh phng G gii hn bi y= 4-x2; y=x2+2 .Quay hnh phng (G) quanh Ox ta-c mt vt th. Tnh th tch vt th ny 6) (HVQY 1997): Cho hnh phng gii hn bi D quay quanh trc Oxy x2; y x Tnh th tch vt th trn xoay khi D

7) (HVKTQS 1995) Tnh th tch do D quay quanh Ox D

;x 2 8) Tnh th tch ca vt th trn xoay sinh ra bi php quay quanh Ox ca hnh phng S gii hn bi cc -ng y=x.ex , x=1 , y=0 (0 x 1 ) ( x 4) 2 y 2 9) (HXD 1998) Tnh th tch vt th to bi hnh ( E ) : 1 quay quanh trc Oy 4 16 1 x2 ;y 10) (HNN1 1999): Cho hnh phng gii hn bi D y 2 x2 1

y

0; y

1 cos4 x sin 4 x ; x

a) b) 11) a) b) 12) a) b) c) 13)

Tnh din tch hnh phng gii hn bi D Tnh th tch vt trn xoay khi D quay quanh Ox (HKT 1996) : Cho hnh phng gii hn bi D y 2 (4 x) 3 ; y 2 4 x Tnh din tch hnh phng gii hn bi D Tnh th tch vt trn xoay khi D quay quanh Ox (HPCCC 2000): Cho hm s (C ) : y x.( x 1) 2 Kho st v v th hm s Vit ph-ng trnh tip tuyn k t 0(0,0) n (C) Tnh th tch gii hn bi (C) quay quanh Ox Cho min (H) gii hn bi -ng cong y=sinx v on 0 x ca trc Ox . Tnh th tch khi trn xoay khi (H) quay quanh a) Trc Ox b) Trc Oy

Chuyn 6:

i s t hp - Nh thc newtn

1. Mt s Bi ton p dng quy tc nhn, cng, hon v, t hp, chnh hp1.1 Cc bi ton chn s: * V d 1: T cc ch s 0,1,2,3,4,5,6 c th lp -c:Bao nhiu s t nhin gm 5 ch s khc nhau. Bao nhiu s t nhin chn gm 5 ch s khc nhau. Bao nhiu s t nhin gm 5 ch s khc nhau trong phi c mt ca s 5. * V d 2: Vi cc ch s 0,1,2,3,4,5 c th lp -c bao nhiu s t nhin tho: a/ Gm 8 ch s t cc s trn. b/ Gm 8 ch s trong ch s 1 c mt 3 ln cn cc ch s khc c mt ng 1 ln. * V d 3: Vi cc ch s 1,2,3,4,5 c th lp -c bao nhiu s t nhin gm 5 ch s khc nhau, trong c hai ch s 1 v 2 khng ng cnh nhau. * V d 4:T 10 ch s 0,1,2,3,4,5,6,7,8,9 c th lp -c bao nhiu s gm 6 ch s khc nhau sao cho : a/ S chia ht cho 5. b/ Trong cc ch s c mt ca ch s 0 v 1. c/ Nh hn 600000. * V d 5: Xt cc hon v ca 6 ch s 1,2,3,4,5,6. Tnh tng S ca tt c cc s to thnh bi cc hon v ny. * V d 6: T cc ch s 1,2,3,4,5,6 c th lp -c bao nhiu s gm 6 ch s khc nhau v trong tng ca 3 ch s u nh hn tng ca 3 ch s cui 1 n v. Convert by TVDT 15 a/ b/ c/

www.luyenthi24h.com www.luyenthi24h.comBi tp

* Bi 1: T cc ch s 1,2,5,6,7,8 c th lp -c bao nhiu s gm 3 ch s khc nhau t 5 ch s trn saocho: a/ S to thnh l mt s chn. b/ S to thnh khng c mt ca ch s 7. c/ S to thnh phi c mt ca ch s 1 v 5. d/ S to thnh nh hn 278. *Bi 2: Cho 8 ch s 0,1,2,3,4,5,6,7. a/ C bao nhiu s t nhin gm 4 ch s khc nhau. b/ C bao nhiu s t nhin chn gm 4 ch s khc nhau. c/ C bao nhiu s t nhin chia ht cho 3 gm 4 ch s khc nhau . *Bi 3: Cho tp A 1, 2,3, 4,5,6,7,8 a/ C bao nhiu tp con X ca A tho iu kin cha 1 v khng cha 2. b/ C bao nhiu s t nhin chn gm 5 ch s khc nhau ly t tp A v khng bt u bi s 123. *Bi 4: Cho tp A 0,1, 2,3, 4,5,6,7 c th lp -c bao nhiu s gm 5 ch s khc nhau ly t tp A sao cho: a/ S to thnh l mt s chn. b/ Mt trong 3 ch s u tin phi bng 1. *Bi 5: Xt nhng s gm 9 ch s, trong c 5 ch s 1 v 4 ch s cn li chn t 2,3,4,5. Hi c bao nhiu s nh- vy nu a/ 5 ch s 1 xp k nhau. b/ Cc ch s -c xp tu . *Bi 6: Cho 7 ch s 0,2,4,5,6,8,9. a/ C bao nhiu s c 3 ch s khc nhau lp t cc s trn. b/ C bao nhiu s c 4 ch s khc nhau, trong nht thit phi c ch s 5. *Bi 7: T 10 ch s 0,1,2,3,4,5,6,7,8,9 c th lp -c bao nhiu s gm 7 ch s a1a 2 ...a 7 tho cc iu kin ch s a 3 l s chn , a 7 khng chia ht cho 5, cc ch s a 4 ;a 5 ;a 6 i mt khc nhau. *Bi 8: Vi cc ch s 0,1,2,3,4,5 ta c th lp -c bao nhiu s : a/ Gm 8 ch s, trong ch s 1 c mt 3 ln, cc ch s khc c mt 1 ln. b/ Gm 6 ch s khc nhau v ch s 2 ng cnh ch s 3. *Bi 9: Ta vit cc s c 6 ch s bng cc ch s 1,2,3,4,5 . Trong mi s -c vit c mt ch s -c xut hin 2 ln cn cc ch s cn li xut hin 1 ln. Hi c bao nhiu s nh- vy. * Bi 10: Cho 7 ch s 1,2,3,4,5,6,7. Xt tp E gm 7 ch s khc nhau vit t cc ch s cho. Chng minh rng tng S ca tt c cc s ca tp E chia ht cho 9.

1.2 Cc bi ton chn cc i t-ng thc t: Dng 1: Tm s cch chn cc i t-ng tho iu kin cho tr-c. * V d 1: C 3 bng hng vng, 3 bng hng trng v 4 bng hng ( cc bng hoa xem nh- i 1 khcnhau) ng-i ta mun chn ra mt b hoa gm 7 bng. a/ C bao nhiu cch chn cc bng hoa -c chn tu . b/ C bao nhiu cch chn sao cho c ng 1 bng mu . c/ C bao nhiu cch chn sao cho c t nht 3 bng hng vng v t nht 3 bng hng . * V d 2: Mt cuc khiu v c 10 nam v 6 n, ng-i ta chn c th t 3 nam v 3 n ghp thnh 3 cp. Hi c bao nhiu cch chn. * V d 3: Mt lp hc c 30 hc sinh trong c 3 cn s lp.n chn 3 em trong 30 hc sinh trn i trc tun sao cho trong 3 em -c chn lun c 1 cn s lp. Hi c bao nhiu cch chn. * V d 4:Mt tr-ng tiu hc c 50 hc sinh tin tin, trong c 4 cp anh em sinh i. Ng-i ta cn chn 3 hc sinh trong 50 hc sinh trn i d hi tri cp thnh ph sao cho khng c cp anh em sinh i no -c chn. Hi c bao nhiu cch chn. * V d 5:Trong mt mn hc, gio vin c 30 cu hi khc nhau gm 5 cu kh , 10 cu trung bnh v 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 (kh, trung bnh v d) ng thi s cu d khng t hn 2. Convert by TVDT 16

www.luyenthi24h.com www.luyenthi24h.com* V d 6: Trong mt phng cho a gic u H c 20 cnh. Xt cc tam gic c 3 nh -c ly t cc nhca H. a/ b/ c/ d/ C bao nhiu tam gic nh- vy. C bao nhiu tam gic c ng 2 cnh l cnh ca H. C bao nhiu tam gic c ng 1 cnh l cnh ca H. C bao nhiu tam gic khng c cnh no l cnh ca H.

Dng 2: Xp v tr cc i t-ng tho iu kin cho tr-c. * V d 7: C bao nhiu cch xp 5 bn A,B,C,D,E vo mt gh di sao choBn C ngi chnh gia. Bn A v E ngi hai u gh. * V d 8: Trong mt phng hc c 2 dy bn di, mi dy c 5 ch ngi. Ng-i ta mun xp ch ngi cho 10 hc sinh gm 5 nam v 5 n. Hi c bao nhiu cch xp nu: a/ Cc hc sinh ngi tu . b/ Cc hc sinh nam ngi mt bn v n ngi mt bn. * V d 9: Mt hi ngh bn trn c 4 phi on cc n-c : Vit Nam 3 ng-i, Lo 5 ng-i, Thi Lan 3 ng-i v Trung Quc 4 ng-i. Hi c bao nhiu cch xp ch ngi cho mi thnh vin sao cho ng-i cng quc tch th ngi gn nhau. * V d 10: Mt bn di c hai dy gh i din nhau, mi dy gm 4 gh. Ng-i ta mun sp xp ch ngi cho 4 hc sinh tr-ng A v 4 hc sinh tr-ng B vo bn ni trn . Hi c bao nhiu cch xp trong mi tr-ng hp sau: a/ Bt c hai hc sinh no ngi cnh nhau hoc i din nhau cng khc tr-ng vi nhau. b/ Bt c hai hc sinh no ngi i din nhau cng khc tr-ng vi nhau. Bi tp * Bi 1: Mt lp hc c 40 hc sinh gm 25 nam v 15 n. C bao nhiu cch chn 4 hc sinh sao cho : a/ S hc sinh nam hoc n l tu . b/ Phi c 2 nam v 2 n. c/ Phi c t nht 1 n. d/ S hc sinh nam khng v-t qu 2. * Bi 2: Mt lp hc c 40 hc sinh cn c ra 1 ban cn s gm 1 lp tr-ng, 1 lp ph v 3 u vin . Hi c my cch lp ra ban cn s lp. * Bi 3: Gia nh ng A c 11 ng-i bn trong c 1 cp v chng. ng mun mi 5 ng-i n d tic, trong c cp v chng c th cng -c mi hoc khng cng -c mi. Hi ng A c bao nhiu cch mi. * Bi45:Mt i thanh nin tnh nguyn c 15 ng-i, gm 12 nam v 3 n. Hi c bao nhiu cch phn cng i thanh nin tnh nguyn v gip 3 tnh mn ni , sao cho mi tnh c 4 nam v 1 n. * Bi 5: i tuyn hc sinh gii ca mt tr-ng gm 18 em, trong c 7 hc sinh khi 12, 6 hc sinh khi 11 v 5 hc sinh khi 10. Hi c bao nhiu cch c 8 hc sinh trong i i d tri h sao cho mi khi c t nht mt em -c chn. * Bi 6: Cho hai -ng thng song song. Trn -ng th nht c 10 im phn bit v -ng thng th hai c 20 im phn bit. C bao nhiu tam gic -c to bi cc im cho. * Bi 7: Cho a gic u A1A2 ...A2n (n 2,n ) ni tip -ng trn tm O. Bit rng s cc tam gic c cc nh l 3 trong 2n im A1;A2 ;...;A2n nhiu gp 20 ln s cc hnh ch nht c cc nh l 4 trong 2n im A1;A2 ;...;A2n . Hy tm n. *Bi 8 : Mt t gm 6 hc sinh A,B,C,D,E,F -c xp vo 6 ch ngi -c ghi s th t trn mt bn di. Tm s cch xp cc hc sinh ny sao cho: a/ A v B ngi chnh gia cc hc sinh cn li. b/ A v B khng ngi cnh nhau. *Bi 9 : Mt hc sinh c 12 cun sch i mt khc nhau trong c 2 cun sch mn ton, 4 cun mn vn, 6 cun mn anh vn. Hi c bao nhiu cch xp tt c cc cun sch ln mt k di , nu mi cun sch ny -c xp k nhau v nhng cun cng mn hc xp k nhau. a/ b/

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17

www.luyenthi24h.com www.luyenthi24h.com* Bi 10: Mt bn di c hai dy gh i din nhau, mi dy gm 6 gh. Ng-i ta mun sp xp ch ngi cho6 hc sinh tr-ng A v 6 hc sinh tr-ng B vo bn ni trn . Hi c bao nhiu cch xp trong mi tr-ng hp sau: a/ Bt c hai hc sinh no ngi cnh nhau hoc i din nhau cng khc tr-ng vi nhau. b/ Bt c hai hc sinh no ngi i din nhau cng khc tr-ng vi nhau.

2. Cc bi ton nh thc, ph-ng trnh bt ph-ng trnh Hon v, t hp & chnh hpMt s kin thc cn nh1. Hon v : Pn

n. n 1 ...2.1n n 1 ... n k 11 ,0 kk k Cn .a n k .b k k 0 k 0

k 2. Chnh hp: An

n! n k !nk Cn

0 O ! 1, An

1 0 kk Cn k Cn

n1

k 3. T hp: Cn

n! O Cn k !. n k !n

n Cn

k

k Cn 1

4. Nh Thc nu tn: a b

n k Cn .a k .b n k

Tng c n+1 s hng .bc ca mi s hng l n-k+k=n k S hng tng qut Tk 1 Cn .a n k .b k

Cc v dI. Gii pt, h pt, bt ph-ng trnh, h bt ph-ng trnh v i s t hp 3 *V d 1. Gii phng trnh: a, C1 6.Cx2 6.Cx 9 x 2 14 x b, C5x 2 C5x 1 C5x 25 x 5 2 14 *V d 2. Gii phng trnh: x C5 C6x C7x *V d 3. Hy tm s nguyn dong tha m phng trnh 5 2 4 3 a, Cn 1 Cn 1 S: n=11 An 2 0 4 2 n 2 3 3 n 0 1 2 n b, Cn .Cn 2 2Cn Cn Cn Cn 3 100 c, Cn 2Cn 4Cn ... 2n Cn 243

*V d 4. Px Ax2 72 6 Ax2 2 Px *V d 5. Gii h phng trnh2 Cn 1 *V d 6. Gii bpt: a) 2 Cn

2 Axy 5Ay x

5C xy 2Cy x

90 80

S: x=5 ,y=2 S: a)

3 n 10

3 n 1 b) An 1 Cn 1 14 n 1

2 3

n 5 b)

7 2

n 4

4 4 An 4 An 143 24 *V d 7. Gii bt phng trnh: a) b) 3 n 4 n 2 ! 4 Pn 23 An 1 Cn S: a) 9,5 n 2,5 b)1 n 5 5 2 1 2 3 2 *V d 8. Gii bt phng trnh: a, Cx4 1 Cx 1 Ax 2 0 b, A2 x Ax 4 2 S: a, 5 x 11 b, x 4 Bi tp 1. Gii cc ph-ng trnh sau:

6 3 Cx 10 x

1/

2A 2 x

50

A2 x 2

2/

1 Cx 4

1 Cx 5

1 Cx 618

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www.luyenthi24h.com www.luyenthi24h.comk k k 2. Tm k sao cho cc s C7 ;C7 1;C7 3. Gii cc bt ph-ng trnh sau: 2

theo th t lp thnh mt cp s cng. 2/

1/

C4 n

1

C3 n

1

5 2 An 4

2

0, n

A3 n

2Cn n

2

9n

4. Gii cc h ph-ng trnh sau: 1/

2A y x 5Ay x

y 5C x

90 806(A 2 x 2Px )3

2C72 2C 2 n

y x

2/

Cy 1 : Cy 1 : Cy 1 x x x

6:5:2

5. Gii cc ph-ng trnh sau: 1/ 3/

Px A 2 x C2 n1

2/4

1 Cx 5

2 Cx 6

14 Cx 7

2

2C 2 n

C2 n

149

4/

C1 x

6C2 x5 2 Ax 4

6C3 x

9x 2 14x

6. Gii cc bt ph-ng trnh sau: Cx 3 1 x 1 1/ 4 A x 1 14P 3 1 2 6 3 A2 x A2 C x 10 3/ x 2 x 7. Gii cc PT v h PT sau: Cy Cy 1 0 x x1/

2/ 4/

C 4 1 C3 1 x x

2

0

4 2x C2 C2x ... C2x 2x

22003 1

4C

y x

5C

y 1 x

02 3

2/

C m 11 : C m 1 : C m 11 n n n

5:5:3

Pn 5 k 60 An (n k )! 9. Gii h ph-ng trnh C xy 1 : C xy 1 : C xy 18. Gii bt ph-ng trnh2 10. Gii bt ph-ng trnh C 2 x 4 C2 x

vi 2 n n, k thuc N (TNPT 2003 - 2004)6:5:2

(TNPT 2002 - 2003)n 2.C n 2

2 ...... C 2 xx

2 2003 1

3 11. Tm s n nguyn d-ng tho mn bt ph-ng trnh An

9n

S: n = 4, n = 32005

12. Tm s t nhin n tho mn: C .C1 2n 1

2 n

n 2 n

2C .C2 2n 1

2 n

3 n2

C .C

3 n

n 3 n

100 .1 2n 1 ...( 2n 1).22 n C2 n 1

2.2C Tm s t nhin n bit (KA 2005) C II. Tm 1 s hng hoc h s ca mt s hng4

3.2 C

3 2n 1

4 4.23 C2 n

1 *V d 1.Tm h s ca s hng cha x trong khai trin x x

10

*V d 2. Tm s hng x31, Trong khai trin x

1 x2

40

*V d 3. Tm s hng khng cha x trong khai trin *V d 4. Trong khai trin x 3 x x28 15 n

3

x

1 4 x

7

n n Tm s hng khng cha x bit Cn Cn

1

n Cn

2

79

*V d 5. Tm h s ca s hng cha x trong khai trin

43

x

5 3

1 x2

21

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19

www.luyenthi24h.com www.luyenthi24h.com1 *V d 6. Bit trong khai trin x 3 ng gia trong khai trinn

C h s ca s hng th 3 bng 5. Hy tnh s hngn

*V d 7. Cho khai trins ca s hng c cha x5

x

3 3

3 x2

. Bit tng ca ba s hng u itn trong khai trin bng 631. Tm h

*V d 8. Bit tng h s ca ba s hng u tin trong khai trinkhng cha x10

x x

3

115

n

x 28

bng 79 .Tm s hng

*V d 9. tm h s ca x y trong khai trin6 2

xy

x y

*V d 10. Trong khai trin .nguyn dng.

3

xy 2

12

xy

. Tm s hng cha x v y sao cho s m ca x v y l cc s19

*V d 11. Tm cc hng t l s nguyn trong khai trin *V d 12.a, Cho khai trin 1 x101

3

3

2

. Trong cc h s ca cc s hng .Tm h s ln nht30

b, Cho khai trin . 1 2x

.Tm h s ln nht trong cc h s

Bi tpa0 a1 x ... a100 x 1. Bit rng (2 x) a) CMR: a2 < a3 . b) Vi gi tr no ca k th ak< ak + 1 (0k99) k 2. Tm k thuc {0, 1, . 2005} sao cho: C 2005 t GTLN.100 100

3. Tm s nguyn n>1 tho mn ng thc: 2 Pn 4. Tnh gi tr ca biu th-c MA4 n1

2 6 An

2 Pn An

12 .

3A 3 n n l s nguyn d-ng Bit rng: (n 1)! 2 2 2 2 C n 1 2C n 2 2C n 3 C n 4 149 5. Tm h s ca x7 trong khai trin thnh a thc ca (2 - 3x) 2n. 6. Gi s (1 2 x) n a0 a1 x ... a n x n v a0 a1 ... an 729.

Tm n v s ln nht trong cc s: a0 , a1 ,..., an 7. Gi s n l s nguyn d-ng v (1 x) n Bit rng k nguyn (0

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