Chuyen de Daiso on Thi Lop10

.:: CHUYN TON LUYN THI VO LP 10 THPT ::.

Bin son: Trn Trung Chnh Trang s 1



KHI QUT KIN THC TP HP

1. Tp hp s t nhin K hiu l: N.

Phn t ca tp hp: N = { 0, 1, 2,, n,}

Cc k hiu khc: Tp hp s t nhin c s "0":

N0 = { 0, 1, 2, ..., n, ...}

Tp hp s t nhin khng cha s "0" l: N

* = {1, 2, ..., n, ...}.

Cc tnh cht ca php cng cc s t nhin: Vi a, b, c l cc s t nhin, ta c: (1) Tnh cht giao hon: a + b = b + a (2) Tnh cht kt hp: (a + b) + c = a + (b + c) (3) Tnh ng nht khi cng: a + 0 = 0 + a = a. (4) Tnh cht phn phi ca php cng i vi php nhn: (b + c)a = b.a + c.a

Cc tnh cht ca php nhn cc s t nhin: Vi a, b, c l cc s t nhin, ta c: (1) Tnh cht giao hon: a.b = b.a (2) Tnh cht kt hp: (a.b).c = a.(b.c) = a.b.c (3) Tnh ng nht khi nhn: a.1 = 1.a = a (4) Tnh cht phn phi ca php nhn i vi php cng: a(b + c) = a.b + a.c

2. Tp hp s nguyn K hiu l: Z.

Phn t ca tp hp:

Z = {0, 1, 2, ..., n, ...} Cc k hiu khc:

Tp hp cc s nguyn m l - N = {-1, -2, ..., -n, ...} Tp hp cc s nguyn dng l + N = {+1, +2, ..., +n, ...} Cc php ton trn s nguyn:

Ton Cng Ton Tr Ton Nhn Ton Chia

a + 0 = a

a + a = 2a

a + (-a) = 0

a - 0 = a

a - a = 0

(a) - (-a) = 2a

a x 0 = 0

a x 1 = a

a x a = a2

a x

a

1= 1

0

a=

1

a = a

a

a= 1

1

a

= -a

3. Tp hp s hu t K hiu l: Q.

Phn t ca tp hp:

mx | x , n 0; m,n

n

ZQ

Mt s k hiu khc: Tp hp cc s hu t khng m l Q+. Tp hp cc s hu t dng l Q*.

Cc cch biu din s hu t: Biu din trong h thp phn v cc h c s khc.

S hu t c th l s thp phn hu hn hoc s thp phn v hn tun hon. Dy cc ch s lp li trong biu din thp phn ca cc s thp phn v hn tun hon c gi l chu k. S cc ch s trong chu k ny c th chng minh c rng khng vt qua gi tr tuyt i ca b.

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Biu din bng lin phn s. Mt s thc l s hu t khi v ch khi biu din lin phn s ca n l hu hn. 4. Tp hp s thc

K hiu l: R Cc k hiu khc:

Tp hp s thc khng m l R+ Tp hp cc s thc dng l R*

Cc php ton trn tp s thc: Php cng, php tr, php nhn, php chia, php ly tha, php logarit.

5. Tp hp s v t K hiu l: I Phn t ca tp hp:

I = R\Q

Trong ton hc, s v t l s thc khng phi l s hu t, ngha l khng th biu din c

di dng t s a

b

, vi a, b l cc s nguyn.

V d: S thp phn v hn c chu k thay i: 0.1010010001000010000010000001...

S 2 = 1,41421 35623 73095 04880 16887 24209 7... S pi () = 3,14159 26535 89793 23846 26433 83279 50288 41971 69399 37510 58209 74944 59230 78164 06286 20899 86280 34825 34211 70679...

6. Cc php ton trn tp hp:

a. Hp ca cc tp hp: nh ngha: Hp ca tp hp A v tp hp B l tp hp gm tt c cc phn t thuc t nht mt trong hai tp hp A v B.

K hiu: A B

Phn t ca A B = {x| x A hoc x B}

b. Giao ca cc tp hp: nh ngha: Giao ca tp hp A v tp hp B l tp hp gm tt c cc phn t thuc hai tp hp A v B.

K hiu: A B

Phn t ca A B = {x| x A v x B}

c. Hiu ca cc tp hp: nh ngha: Hiu ca tp hp A v tp hp B l tp hp gm tt c cc phn t thuc tp hp A nhng khng thuc tp hp B.

K hiu: A \ B

Phn t ca A \ B = {x| x A v x B}

d. Phn ca cc tp hp:

nh ngha: Nu A B th B\A c gi l phn b ca tp hp A trong tp hp B. K hiu: CAB.



CHUYN 2 CN THC

1. Cn bc hai: Khi nim: x c gi l cn bc hai ca s khng m a

x2 = a.

K hiu: x a , vi a 0.

iu kin xc nh ca biu thc A l: A xc nh A 0 . V d:

(1) Cn bc hai ca 25 l 525 .

(2) Cn bc hai ca 12 l 3212 .

(3) iu kin 2x c ngha l x - 2 0 x 2.

(4) Tnh 23x .

Ta c: 3x3x3x 2 Hay

3x3x 2 v 3x3x3x 2 Cc php bin i cn thc

A.B A. B, A 0; B 0

A A

, A 0; B 0B B

2A B A B, B 0

A 1

A.B, A.B 0; B 0B B

22m. A Bm

, B 0; A BA BA B

n. A Bn

, A 0; B 0; A BA BA B

2

A 2 B m 2 m.n n m n m n , (vi m, n 0, vi

m n A

m.n B

2

AA Lu : Vi mi s thc a, gi tr tuyt i ca a.

K hiu: |a| nh ngha:

a a 0

a a a < 0

nu

nu

nh ngha trn cho thy, gi tr tuyt i ca a l mt s khng m. 2. Cn bc ba:

K hiu: Cn bc ba ca mt biu thc (hoc mt s) A l: 3 A .

Ta c: 3 3A A . V d:

1) 3 33 8 2 2 2)

33 x 2 x 2

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3. Cn bc cao: Cn bc chn: Vi mi s t nhin m, n, k > 1, ta c:

2k 2kA A .

2k 2k 2kA.B A . B , A.B 0

2k

2k

2k

AA, A.B 0; B 0

B B

2k2k 2kA .B A . B, B 0

m m.nn A A, A 0 V d:

(1) Cn bc 4 ca 16 l 4 44 16 2 2.

(2) Cn bc 4 ca (x + 2)2 l 2

4 x 2 x 2 , (x + 2 0).

Ch : 2k A c ngha khi A 0. Cn bc l: Vi mi s t nhin m, n, k > 1, ta c:

2k 1 2k 1A A.

2k 1 2k 12k 1 A.B A. B

2k 1

2k 12k 1

A A, B 0

B B

2k 12k 1 2k 1A .B A. B

V d:

(1) Cn bc 3 ca 27 l 3273 .

(2) Cn bc 3 ca (4 - x)3 l x4x43 3 . Ch : i vi cn bc l th biu thc trong du cn khng quy nh du m hoc dng.



CHUYN 3 HNG NG THC

1. Kin thc c bn: 1.1. hng ng thc ng nh: (a + b)

2 = a

2 + 2ab + b

2 (Bnh phng ca mt tng)

(a - b)2 = a

2 - 2ab + b

2 (Bnh phng ca mt tng)

a2 - b

2 = (a - b)(a + b) (Hiu hai bnh phng)

(a + b)3 = a

3 + 3a

2b + 3ab

2 + b

3 (Lp phng ca mt tng)

(a - b)3 = a

3 - 3a

2b + 3ab

2 - b

3 (Lp phng ca mt tng)

a3 + b

3 = (a + b)(a

2 - ab + b

2) (Tng hai lp phng) a

3 - b

3 = (a - b)(a

2 + ab + b

2) (Hiu hai lp phng) 1.2. Cc hng ng thc nng cao: (a + b + c)

2 = a

2 + b

2 + c

2 + 2ab + 2bc + 2ac

(a + b + c)3 = a

3 + b

3 + c

3 + 3(a + b)(b + c)(c + a)

a3 + b

3 + c

3 - 3abc = (a + b + c)(a

2 + b

2 + c

2 - ab -bc - ca)

an - b

n = (a - b)(a

n-1 + a

n-2b + +abn-1 + bn-1)

(a + b)n = k k n-k

nC a b

= 0 n 1 n-1 2 n-2 2 k n-k k n-1 n-1 n nn n n n n n

C a + C a b + C a b +...+ C a b +...+ C ab + C b (Nh thc

Newton)

(Vi

k

n

n!C =

k! n - k ! v n! = 1.2.3.4(n-1).n)

Ch : n! c l n giai tha.

2. Bi tp p dng: Bi tp 1: Phn tch hng ng thc sau: a) (3 - 2x)

2 b) (2x + 1)

2 c) 9 - 25x

2

Gii a) (3 - 2x)

2 = 3

2 - 2.3.2x + (2x)

2 = 9 - 12x + 4x

2

b) (2x + 1)2 = (2x)

2 + 2.2x.1 + 1

2 = 4x

2 + 4x + 1

c) 9 - 25x2 = 3

2 - (5x)

2 = (3 + 5x)(3 - 5x)

Bi tp 2: Phn tch hng ng thc sau: a) (7 + 3x)

3 b) (9x + 2)

3

Gii a) (7 + 3x)

3 = 7

3 + 3.7

2.3x + 3.7.(3x)

2 + (3x)

3 = 343 + 441x + 189x

2 + 27x

3

b) (9x - 2)3 = (9x)

3 - 3.(9x)

2.2 + 3.9x.2

2 - 2

3 = 729x

3 - 486x

2 + 108x - 8

Bi tp 3: Phn tch hng ng thc sau: a) 1 - 27x

3 b) 216x

3 + 8

Gii a) 1 - 27x

3 = 1

3 - (3x)

3 = (1 - 3x)[1

2 + 1.3x + (3x)

2] = (1 - 3x)(1+ 3x + 9x

2)

b) 216x3 + 8 = (6x)

3 + 2

3 = (6x + 2)[(6x)

2 - 6x.2 + 2

2] = (6x + 2)(36x

2 - 12x + 4)

Bi tp 4: a v dng hng ng thc: a) 2x

2 + 4x + 2 b) x

2 - 6x + 9

c) x3 + 12x

2 + 48x + 64 d) 8x

3 - 12x

2 + 6x - 1

Gii a) 2x

2 + 4x + 2 = 2(x

2 + 2.x.1 + 1

2) = 2(x + 1)

2

b) x2 - 6x + 9 = x

2 - 2.x.3 + 3

2 = (x - 3)

2

c) x3 + 12x

2 + 48x + 64 = x

3 + 3.x

2.4 + 3.x.4

2 + 4

3 = (x + 4)

3

d) 8x3 - 12x

2 + 6x - 1 = (2x)

3 - 3.(2x)

2.1 + 3.2x.1

2 - 1

3 = (2x - 1)

3

Bi tp 5: Phn tch hng ng thc sau: a) (x

2 + x + 1)

2 b) (x

2 + 2x - 3)

2

Gii a) (x

2 + x + 1)

2 = (x

2)

2 + x

2 + 1

2 + 2.x

2.x + 2.x

2.1 + 2.x.1 = x

4 + x

2 + 1 + 2x

3 + 2x

2 + 2x

= x4 + 2x

3 + 3x

2 + 2x + 1

b) (x2 + 2x - 3)

2 = (x

2)

2 + (2x)

2 + 3

2 + 2.x

2.2x - 2.x

2.3 - 2.2x.3 = x

4 + 4x

2 + 9 + 4x

3 - 6x

2 - 12x

= x4 + 4x

3 - 2x

2 - 12x + 9

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Bi tp 6: Tnh nhanh:

a) 20042 - 16 b) 892

2 + 892.216 + 108

2

c) 993 + 1 + 3(99

2 + 99) d) 20,03.45 + 20,03.47 + 20,03.48

Gii a) 2004

2 - 16 = 2004

2 - 4

2 = (2004 - 4)(2004 + 4) = 2000.2008 = 4016000.

b) 8922 + 892.216 + 108

2 = 892

2 + 2. 892.108 + 108

2 = (892 + 108)

2 = 1000

2 = 1000000.

c) 993 + 1 + 3(99

2 + 99) = 99

3 + 3.99

2 + 3.99 + 1

3 = (99 + 1)

3 = 100

3 = 1000000.

d) 20,03.45 + 20,03.47 + 20,03.48 = 20,03(45 + 47 + 48) = 20,03.200 = 20,03.2.100 = 4006.

Bi tp 7 : Vit biu thc 2

4n 3 25 thnh tch

Gii

2

4n 3 25 = (4n + 3)2 - 52 = (4n + 3 + 5)(4n + 3 - 5) = (4n + 8)(4n - 2)

Bi tp 8 : Chng minh vi mi s nguyn n biu thc 2

2n 3 9 chia ht cho 4.

Gii Ta c: (2n + 3)

2 - 9 = (2n + 3)

2 - 3

2 = (2n + 3 + 3)(2n + 3 - 3) = (2n + 6)2n = 4n(n + 3)

Biu thc 4n(n + 3) lun chia ht cho 4. Vy (2n + 3)2 - 9 chia ht cho 4. Bi tp 9: Vit biu thc sau di dng tch

a) 2 2

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

b) 2 2

x y z y z

c) 2

x 3 4 x 3 4

d) 2

25 10 x 1 x 1

Gii

a) 2 2

x + y + z - 2 x + y + z y + z + y + z = [(x + y + z) - ( y + z)]2

= (x + y + z - y - z)2 = x

2.

b) 2 2

x y z y z = [(x + y + z) + (y + z)][(x + y + z) - ( y + z)]

= (x + y + z + y + z)(x + y + z - y - z)

= x(x + 2y + 2z)

c) 2

x 3 4 x 3 4 = (x + 3)2 + 2.(x + 3).2 + 22

= [(x + 3) + 2]2 = (x + 3 + 2)

2

= (x + 5)2

d) 2

25 10 x 1 x 1 = 52 + 2. 5.(x + 1) + (x + 1)2

= [5 + (x + 1)]2

= (5 + x + 1)2

= (x + 6)2

Bi tp 10: Vit biu thc sau di dng hng ng thc:

a) x y z t . x y z t

b) x y z t . x y z t

2 4c) 2 3 1 3 1 3 1 Gii

a) x y z t . x y z t = [(x + y) + (z + t)][(x + y) - (z - t)] = (x + y)

2 - (z - t)

2

b) x y z t x y z t = [(x - y) + (z - t)] [(x - y) - (z - t)] = (x - y)

2 - (z - t)

2

2 4c) 2 3 1 3 1 3 1 = (3 - 1)(3 + 1)(3

2 + 1)(3

4 + 1)

= (32 - 1)(3

2 + 1)(3

4 + 1)



= (34 - 1)(3

4 + 1)

= 38 - 1

3. Bi tp t luyn: Bi tp 1: Phn tch cc hng ng thc sau: a) (3x + 4)

2 b) (2x - 5)

2 c) 49 - x

4

Bi tp 2: Phn tch cc hng ng thc sau: a) (x + y + z)

3 b) (y - z + t)

3

c) 8x3 - 125 b) 27y

3 + 64z

3

Bi tp 3: Vit cc biu thc sau di dng hng ng thc: a) x

2 - 6x + 9 b) 25 + 10x + x

2

c) x3 + 15x

2 + 75x + 125 d) x

3 - 9x

2 + 27x - 27

Bi tp 4: Vit cc biu thc sau di dng hng ng thc: a) x

2 + 10x + 26 + y

2+ 2y b) x

2 - 2xy + 2y

2 + 2y + 1

c) x2 - 6x + 5 - y

2 - 4y d) 4x

2 - 12x - y

2 + 2y + 1

Bi tp 5: Rt gn biu thc: a) (x + 1)

2 - (x - 1)

2 - 3(x + 1)(x - 1)

b) 5(x - 2)(x + 2) - 21

6 8x 172

c) (x + y)3 + (x - y)

3

d) (x + y - z)2 - (x - z)

3 - 2xy + 2yz.

Bi tp 6: Cho x + y = 7. Tnh gi tr ca biu thc: M = (x + y)3 + 2x2 + 4xy + 2y2. Bi tp 7: Cho x - y = 7. Tnh gi tr ca biu thc: A = x(x + 2) + y(y - 2) - 2xy + 37. Bi tp 8: Cho a2 + b2 + c2 + 3 = 2(a + b + c). Chng minh rng: a = b = c = 1. Bi tp 9: Chng minh rng: (10a + 5)2 = 100a(a + 1) + 25. T hy nu nhng cch tnh nhm bnh phng ca mt s t nhin c tn cng bng ch s 5.

p dng tnh: 252; 352; 652; 752. Bi tp 10: Tnh: A = 12 22 + 32 42 + 20042 + 20052.

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CHUYN 4 PHN TCH A THC THNH NHN T

1. Kin thc cn nh: Phn tch a thc thnh nhn t l mt kin thc thuc chng trnh Ton lp 8. y l dng ton tng i phc tp. Loi ton ny thng c p dng rng ri trong cc k thi HSG, thi chuyn cp, thi vo trng chuyn, ... Cc phng php phn tch a thc thnh nhn t: Phng php 1: Dng hng ng thc ng nh. Phng php 2: t nhn t chung. Phng php 3: Tch hng t. Phng php 4: Phi hp nhiu phng php. Phng php 5: Thm v bt cng mt hng t. Phng php 6: i bin s. Phng php 7: Xt gi tr ring. Phng php 8: Dng h s bt nh. Phng php 9: Nhm nghim.

2. Phng php dng hng ng thc ng nh Phng php: Nm chc 7 hng ng thc ng nh v cc hng ng thc nng cao. Nhn dng hng ng thc vi cc dng biu thc phc tp. V d: Nu ta bit hng ng thc bnh phng ca mt tng l (A + B)2 th [(A + C) + (B - C)]2 ta phi bit. H bc ly tha ca mt bin hoc mt s v a v dng hng ng thc. Thm mt cht t duy, sng to trong cch bin i xut hin hng ng thc.

a) Bi tp p dng: Bi tp 1: Phn tch a thc (x + y)2 (x y)2 thnh nhn t.

Gii (x + y)

2 (x y)2 = [(x + y) (x y)].[(x + y) + (x y)]

= (x + y x + y)(x + y + x y) = 2y.2x = 4xy.

Bi tp 2: Phn tch a6 b6 thnh nhn t. Gii

a6 b6 =

2 23 3a b = (a3 b3 )( a3 + b3 )

= (a b)(a2 + ab + b2)(a + b)(a2 ab + b2) Bi tp 3: Phn tch a thc x12 - y4 thnh nhn t.

Gii x

12 - y

4 = (x

6)

2 - (y

2)

2 = (x

6 + y

2)(x

6 - y

2) = (x

6 + y

2)(x

3 - y)(x

3 + y)

Bi tp 4: Phn tch a thc sau thnh nhn t: x4 - 4x3 + 4x - 1 Gii x

4 - 4x

3 + 4x - 1 = (x

4 - 4x

3 + 4x

2) - (4x

2 - 4x + 1)

= x2(x - 2)

2 - (2x - 1)

2 = [(x(x - 2) + 2x - 1][x(x - 2) - (2x - 1)]

= (x2 - 1)(x

2 - 4x + 1) = (x + 1)(x - 1)(x

2 - 4x + 1)

Bi tp 5: Phn tch a thc sau thnh nhn t: x4 - 2x3 - 3x2 + 4x + 4

Gii x

4 - 2x

3 - 3x

2 + 4x + 4 = (x

2 - 1)

2 - 2(x

2 - 1)(x + 1) + (x + 1)

2

= [(x2 - 1) - (x + 1)]

2 = (x + 1)

2(x - 2)

2

Bi tp 6: Phn tch a thc sau thnh nhn t: 9x2 4

Gii 9x

2 4 = (3x)2 22 = ( 3x 2)(3x + 2)

Bi tp 7: Phn tch a thc sau thnh nhn t: 8 27a3b6 Gii 8 27a3b6 = 23 (3ab2)3 = (2 3ab2)( 4 + 6ab2 + 9a2b4) Bi tp 8: Phn tch a thc sau thnh nhn t: 25x4 10x2y + y2



Gii 25x

4 10x2y + y2 = (5x2 y)2

Bi tp 9: Phn tch a thc sau thnh nhn t: a16 + a8b8 + b16

Gii Ta c th vit: a

16 + a

8b

8 + b

16 = a

16 + 2a

8b

8 + b

16 - a

8b

8

= (a8 + b

8)

2 - (a

4b

4)

2

= (a8 + b

8 - a

4b

4)( (a

8 + b

8 + a

4b

4)

Ta li c: a

8 + b

8 + a

4b

4 = (a

4 + b

4)

2 - (a

2b

2)

2

= (a4 + b

4 - a

2b

2)(a

4 + b

4 + a

2b

2)

Mt khc: a

4 + b

4 + a

2b

2 = (a

2 + b

2)

2 - (ab)

2

= (a2 + b

2 - ab)(a

2 + b

2 + ab)

Do , ta c: a

16 + a

8b

8 + b

16 = (a

8 - a

4b

4 + b

8)(a

4 - a

2b

2 + b

4)(a

2 - ab + b

2)(a

2 + ab + b

2)

Bi tp 10: Phn tch a thc sau ra tha s: A = x4 + 6x3 + 7x2 - 6x + 1

Gii Ta c th vit: A = x

4 + 6x

3 + 7x

2 - 6x + 1

= (x4 + 3x

3 - x

2) + (3x

3 + 9x

2 - 3x) - x

2 - 3x + 1

= x2(x

2 + 3x - 1) + 3x(x

2 + 3x - 1) - (x

2 + 3x - 1)

= (x2 + 3x - 1)

2

Vy A = (x2 + 3x - 1)2

b) Bi tp t luyn: Bi tp 1: Phn tch a thc sau thnh nhn t: (x + y)2 - 9x2 Bi tp 2: Phn tch a thc (2n + 5)2 - 25 thnh nhn t. Bi tp 3: Phn tch a thc thnh nhn t: 64 - 27a3b6. Bi tp 4: Phn tch a thc thnh nhn t: 4(x +1)2 - 25(x - 1)4 Bi tp 5: Phn tch a thc thnh nhn t: 25(2x +3)2 - 10 (4x2 - 9) + (2x - 3)2

Bi tp 6: Phn tch a thc thnh nhn t: x4+ x3 + 2x2 + x +1 Bi tp 7: Phn tch a thc thnh nhn t: x3 + 2x2y + xy2 - 9x Bi tp 8: Phn tch a thc thnh nhn t: (a + b + c)3 - a3 - b3 - c3. Bi tp 9: Phn tch cc a thc thnh nhn t: a) A = (a + 1)(a + 3)(a + 5)(a + 7) + 15

b) B = 4x2y

2(2x + y) + y

2z

2(z - y) - 4z

2x

2(2x + z)

3. Phng php t nhn t chung Phng php: Tm nhn t chung ca cc h s nu c (CLN ca cc h s) hoc l nhng n, a thc c mt trong tt c cc hng t. Phn tch mi hng t thnh tch ca nhn t chung v mt nhn t khc hoc nhn t chung ca cc bin (mi bin chung ly s m nh nht). Nhm a v dng: A.B + A.C = A(B + C).

A.B + A.C + A.D = A.(B + C + D).

Vit nhn t chung ra ngoi du ngoc, vit cc nhn t cn li ca mi hng t vo trong du ngoc (k c du ca chng). Lu : i vi a thc th ta c cch bin i nh sau: Tm nghim ca a thc (i xng th c th l -1 hoc 1) i vi cc a thc bc chn th ta chia cho x2 (vi x2 khng l nghim ca a thc). i vi a thc bc l th ta nhm nghim l thng ca c hng t c s m cao nht v c ca hng t t do. Ri a a thc v a thc bc l v lm tng t. Ta c th p dng thm quy tc ng nht h s (ch phi gii h phng trnh hoc cch khc tm cc h s ca cc a thc): V d: Phn tch a thc: ax2 + bx + c = (ax + d)(x + e)

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Mt a thc bc hai c th phn tch thnh tch ca hai a thc bc nht. Mt a thc bc ba c th phn tch thnh tch ca hai a thc bc nht v bc hai. Cc a thc cn li th c th phn tch tng t. Dng chung:

n n 1 p p 1 q q 1n n 1 1 0 p p 1 1 0 q q 1 1 0a x a x ... a x a a x a x ... a x a a x a x ... a x a (Vi p + q = n v p, q, n N) a) Bi tp p dng: Bi tp 1: Phn tch a thc 14x2 y 21xy2 + 28x2y2 thnh nhn t. Gii 14x

2 y 21xy2 + 28x2y2 = 7xy.2x 7xy.3y + 7xy.4xy

= 7xy.(2x 3y + 4xy) Bi tp 2: Phn tch a thc 10x(x y) 8y(y x) thnh nhn t. Gii 10x(x y) 8y(y x) = 10x(x y) + 8y(x y) = 2(x y).5x + 2(x y).4y = 2(x y)(5x + 4y) Bi tp 3: Phn tch a thc 9x(x y) 10(y x)2 thnh nhn t.

Gii 9x(x y) 10(y x)2 = 9x(x y) 10(x y)2 = (x y)[9x 10(x y)] = (x y)(10y x) Bi tp 4: Phn tch a thc sau thnh nhn t: 2x(y z) + 5y(z y ) Gii 2x(y z) + 5y(z y ) = 2(y - z) 5y(y - z) = (y z)(2 - 5y) Bi tp 5: Phn tch a thc sau thnh nhn t: xm + xm + 3

Gii x

m + x

m + 3 = x

m (x

3 + 1) = x

m( x+ 1)(x

2 x + 1)

b) Bi tp t luyn: Bi tp 1: Phn tch a thc x3 - 2x2 + x thnh nhn t.

Bi tp 2: Phn tch a thc 2 3 3 4 3 35x y 25x y 10x y thnh nhn t.

Bi tp 3: Phn tch a thc 5(x - y) - y(x - y) thnh nhn t.

Bi tp 4: Phn tch a thc 4 2 2 5 415x 10x y 5x y thnh nhn t.

Bi tp 5: Phn tch a thc xt(z - y) - yt(y - z) thnh nhn t.

3. Phng php nhm hng t Phng php: Dng cc tnh cht giao hon, kt hp ca php cng cc a thc, ta kp hp nhng hng t ca a thc thnh tng nhm thch hp, ri dng cc phng php khc phn tch nhn t theo tng nhm v phn tch chung i vi cc nhm. La chn cc hng t thch hp thnh lp nhm nhm lm xut hin mt trong hai dng sau hoc l t nhn t chung, hoc l dng hng ng thc. Thng thng ta da vo cc mi quan h sau: Quan h gia cc h s, gia cc bin ca cc hng t trong bi ton. Thnh lp nhm da theo mi quan h , phi tho mn: Mi nhm u phn tch c. Sau khi phn tch a thc thnh nhn t mi nhm th qu trnh phn tch thnh nhn t phi tip tc thc hin c na. Dng bi ton: A.B + A.C + E.B + E.C = (A.B + A.C) + (E.B + F.C)

= A(B + C) + E(B + C)

= (B + C)(A + E)

a) Bi tp p dng: Bi tp 1: Phn tch a thc x2 xy + x y thnh nhn t. Gii x

2 xy + x y = (x2 xy) + (x y)



= x(x y) + 1.(x y) = (x y)(x + 1) Bi tp 2: Phn tch a thc x2 2x + 1 4y2 thnh nhn t.

Gii x

2 2x + 1 4y2 = (x2 2x + 1) (2y)2

= (x 1)2 (2y)2 = (x 1 2y)(x 1 + 2y) Bi tp 3: Phn tch a thc x2 2x 4y2 4y thnh nhn t. Gii x

2 2x 4y2 4y = (x2 4y2 ) + ( 2x 4y )

= (x + 2y)(x 2y) 2(x + 2y) = (x + 2y)(x 2y 2)

Bi tp 4: Phn tch a thc sau thnh nhn t: P = 4 3 2 2x x 2 m 1 x mx m . Gii

2 2 4 3 2

2 22 2 4 2

2 2

2

2 2

P m m 2x x x x 2x

x x x 9x m 2m x x 2x .

2 2 4 4

x 3x m x

2 2

m x 2x m x x

Bi tp 5: Phn tch a thc sau thnh nhn t: 2x3 3x2 + 2x 3 Gii 2x

3 3x2 + 2x 3 = ( 2x3 + 2x) (3x2 + 3)

= 2x(x2 + 1) 3( x2 + 1)

= ( x2 + 1)( 2x 3)

Bi tp 6: Phn tch a thc sau thnh nhn t: x2 2xy + y2 16

Gii x

2 2xy + y2 16 = (x y)2 - 42

= ( x y 4)( x y + 4) b) Bi tp t luyn Bi tp 1: Phn tch a thc xy + xz + 3y + 3z thnh nhn t. Bi tp 2: Phn tch a thc x3 - 4x2 + 4x - 1 thnh nhn t. Bi tp 3: Phn tch a thc x2y2 + 1 - y2 - x2 thnh nhn t. Bi tp 4: Phn tch a thc a3 + b3 - a - b thnh nhn t. Bi tp 5: Phn tch a thc a3 + a2b - ab2 - b3 thnh nhn t. 4. Phng php tch hng t Phng php: Tch hng t thnh nhiu hng t nhm: Lm xut hin hng ng thc hiu ca hai bnh phng hoc hiu ca hai hng t l an - bn. Lm xut hin cc h s mi hng t t l vi nhau, nh lm xut hin nhn t chung. Lm xut hin hng ng thc v nhn t chung. Vic tch hng t thnh nhiu hng t khc l nhm lm xut hin cc phng php hc nh: t nhn t chung, dng hng ng thc, nhm nhiu hng t l vic lm ht sc cn thit i vi hc sinh trong gii ton. Ch :

phn tch a thc dng tam thc bc hai: ax2 + bx + c, (a 0) thnh nhn t. Ta tch hng t: bx = b1x + b2x sao cho b1b2 = ac i vi a thc f(x) c bc t ba tr ln, lm xut hin cc h s t l, tu theo c im ca cc h s m ta c cch tch ring cho ph hp nhm vn dng phng php nhm hoc hng ng thc hoc t nhn t chung. Phng php chung: Bc 1: Tm tch ac, ri phn tch a.c ra tch ca hai tha s nguyn bng mi cch:

a.c = a1.c1 = a2.c2 = a3.c3 = = ai.ci =

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Bc 2: Chn hai tha s c tng bng b, chng hn chn tch: a.c = ai.ci vi b = ai + ci

Bc 3: Tch bx = aix + cix. T nhm hai s hng thch hp phn tch tip. a) Bi tp p dng: Bi tp 1: Phn tch a thc f(x) = 3x2 8x + 4 thnh nhn t. Gii Cch 1 (tch hng t 3x2) 3x

2 8x + 4 = 4x2 8x + 4 x2

= (2x 2)2 x2 = (2x 2 x)( 2x 2 + x) = (x 2)(3x 2) Cch 2 (tch hng t : 8x) 3x

2 8x + 4 = 3x2 6x 2x + 4

= 3x(x 2) 2(x 2) = (x 2)(3x 2) Cch 3 (tch hng t : 4) 3x

2 8x + 4 = 3x2 12 8x + 16

= 3(x2 22 ) 8(x 2)

= 3(x 2)(x + 2) 8(x 2) = (x 2)(3x + 6 8) = (x 2)(3x 2) Bi tp 2: Phn tch a thc 6x2 + 7x 2 thnh nhn t. Gii 6x2 + 7x 2 = 6x2 + 4x + 3x 2 = ( 6x2 + 4x) + (3x 2) = 2x(3x 2) + (3x 2) = (3x 2)(2x + 1) Bi tp 3: Phn tch a thc sau ra tha s: n3 7n + 6 Gii n

3 7n + 6 = n3 n 6n + 6

= n(n2 1) 6(n 1)

= n(n 1)(n + 1) 6(n 1) = (n 1)[n(n + 1) 6] = (n 1)(n2 + n 6) = (n 1)(n2 2n + 3n 6) = (n 1)(n(n 2) + 3(n 2)) = (n 1)(n 2)(n + 3) Bi tp 4: Phn tch a thc x4 30x2 + 31x 30 thnh nhn t. Gii Ta c cch tch nh sau: x

4 30x2 + 31x 30 = x4 + x 30x2 + 30x 30

= x(x3 + 1) 30(x2 x + 1)

= x(x + 1)(x2 x + 1) 30(x2 x + 1)

= (x2 x + 1)(x2 + x 30)

= (x2 x + 1)(x 5)(x + 6)

Bi tp 5: Phn tch a thc A = 9x2 - 10x + 1 thnh nhn t. Gii Cch 1: Tch hng t "bc nht", lm xut hin hai tch c hai nhn t chung: A = 9x

2 - 9x - x + 1 = (9x

2 - 9x) - (x - 1)

= 9x(x - 1) - (x - 1) = (x - 1)(9x - 1)

Cch 2: Tch hng t bc hai thnh: A = 10x

2 - 10x - x

2 + 1

= (10x2 - 10x) - (x

2 - 1) = 10x(x - 1) - (x + 1)(x - 1)

= (x - 1)[10x - (x + 1)] = (x - 1)(9x - 1).

Bi tp 6: Phn tch a thc A = x4 + x2 + 1 thnh nhn t:



Gii A = x

4 + 2x

2 + 1 - x

2 = (x

2 + 1)

2 - x

2 = (x

2 - x + 1)(x

2 + x + 1)

Bi tp 7: Phn tch a thc A = 2(x2 + x - 5)2 - 5(x2 + x) + 28 thnh nhn t:

Gii A = 2(x

2 + x - 5)

2 - 5(x

2 + x - 5) + 3

= 2(x2 + x - 5)

2 - 2(x

2 + x - 5) - 3(x

2 + x - 5) + 3

= [2(x2 + x - 5)

2 - 2(x

2 + x - 5)] - [3(x

2 + x - 5) - 3]

= 2(x2 + x - 5)[ (x

2 + x - 5) - 1] - 3[(x

2 + x - 5) - 1]

= (x2 + x - 6)(2x

2 + 2x - 13) = (x - 2)(x + 3)( 2x

2 + 2x - 13)

Ch : Ta c th t n ph: y = x2 + x - 5. Khi A = 2y2 - 5y + 3 Bi tp 8: Phn tch a thc f(x) = 3x2 + 8x + 4 thnh nhn t. Hng dn Phn tch ac = 12 = 3.4 = (3).(4) = 2.6 = (2).(6) = 1.12 = (1).(12) Tch ca hai tha s c tng bng b = 8 l tch a.c = 2.6 (a.c = ai.ci). Tch 8x = 2x + 6x (bx = aix + cix)

Gii Cch 1: Tch hng t bx. 3x

2 + 8x + 4 = 3x

2 + 2x + 6x + 4 = (3x

2 + 2x) + (6x + 4)

= x(3x + 2) + 2(3x + 2)

= (x + 2)(3x +2)

Cch 2 (tch hng t bc hai ax2) Lm xut hin hiu hai bnh phng: f(x) = (4x

2 + 8x + 4) x2 = (2x + 2)2 x2 = (2x + 2 x)(2x + 2 + x)

= (x + 2)(3x + 2)

Tch thnh 4 s hng ri nhm : f(x) = 4x

2 x2 + 8x + 4 = (4x2 + 8x) ( x2 4) = 4x(x + 2) (x 2)(x + 2)

= (x + 2)(3x + 2)

f(x) = (12x2 + 8x) (9x2 4) = = (x + 2)(3x + 2)

Cch 3 (tch hng t t do "c") Tch thnh 4 s hng ri nhm thnh hai nhm: f(x) = 3x

2 + 8x + 16 12 = (3x2 12) + (8x + 16) = = (x + 2)(3x + 2)

Cch 4 (tch 2 s hng, 3 s hng) f(x) = (3x

2 + 12x + 12) (4x + 8) = 3(x + 2)2 4(x + 2) = (x + 2)(3x 2)

f(x) = (x2 + 4x + 4) + (2x

2 + 4x) = = (x + 2)(3x + 2)

Cch 5 (nhm nghim) Ch : Nu f(x) = ax2 + bx + c c dng A2 2AB + C th ta tch nh sau:

f(x) = A2 2AB + B

2 B2 + C = (A B)2 (B2 C)

Bi tp 9: Phn tch a thc f(x) = 4x2 - 4x - 3 thnh nhn t. Hng dn Ta thy 4x2 - 4x = (2x)2 - 2.2x. T ta cn thm v bt 12 = 1 xut hin hng ng thc.

Gii f(x) = (4x

2 4x + 1) 4 = (2x 1)2 22 = (2x 3)(2x + 1)

Bi tp 10: Phn tch a thc f(x) = 9x2 + 12x 5 thnh nhn t. Gii Cch 1: f(x) = 9x

2 3x + 15x 5 = (9x2 3x) + (15x 5)

= 3x(3x 1) + 5(3x 1) = (3x 1)(3x + 5) Cch 2: f(x) = (9x

2 + 12x + 4) 9 = (3x + 2)2 32 = (3x 1)(3x + 5)

Bi tp 11: Phn tch a thc sau thnh nhn t: 2x2 - 5xy + 2y2. Gii Phn tich a thc nay tng t nh phn tich a thc : f(x) = ax2 + bx + c. Ta tach hang t th 2 : 2x

2 - 5xy + 2y

2 = (2x

2 - 4xy) - (xy - 2y

2) = 2x(x - 2y) - y(x - 2y).

b) Bi tp t luyn: Bi tp 1: Phn tch a thc x2 - 5x + 6 thnh nhn t. Bi tp 2: Phn tch a thc 3x2 - 16x + 5 thnh nhn t.

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Bi tp 3: Phn tch a thc 8x2 + 30x + 7 thnh nhn t. Bi tp 4: Phn tch a thc 6x2 - 7x - 20 thnh nhn t. Bi tp 5: Phn tch a thc x3 - 5x2 + 8x - 4 thnh nhn t.

5. Phng php phi hp nhiu phng php Phng php: L s kt hp gia cc phng php nhm nhiu hng t, t nhn t chung, dng hng ng thc, tch hng t. Bit k thut nhn bit dng bit cch p dng phng php no. a) Bi tp p dng: Bi tp 1: Phn tch a thc x4 9x3 + x2 9x thnh nhn t. Gii x

4 9x3 + x2 9x = x(x3 9x2 + x 9)

= x[(x3 9x2 ) + (x 9)]

= x[x2 (x 9) + 1.(x 9)]

= x(x 9)(x2 + 1) Bi tp 2: Phn tch a thc A = (x + y + z)3 x3 y3 z3 thnh nhn t. Gii A = (x + y + z)

3 x3 y3 z3 = [(x + y) + z]3 x3 y3 z3

= (x + y)3 + z

3 + 3z(x + y)(x + y + z) x3 y3 z3

= [(x + y)3 x3 y3 ] + 3z(x + y)(x + y + z)

= 3xy(x + y) + 3(x + y)(xz + yz + z2 )

= 3(x + y)( xy + xz + yz + z2)

= 3(x + y)(y + z)(x + z)

Bi tp 3: Phn tch a thc A = (a - b)3 + (b - c)3 + (c - a)3 thnh nhn t:

Gii Ch : Nu: m + n + p = 0 th m3 + n3 + p3 = 3mnp. Nhn thy: (a - b) + (b - c) + (c - a) = 0 nn ta c ngay: A = (a - b)

3 + (b - c)

3 + (c - a)

3 = 3(a - b)(b - c)(c - a).

Bi tp 4: Phn tch cac a thc sau thnh nhn t : 3xy2 12xy + 12x Gii 3xy

2 12xy + 12x = 3x(y2 4y + 4) = 3x(y 2)2

Bi tp 5: Phn tch cac a thc sau thnh nhn t : 3x

3y 6x2y 3xy3 6axy2 3a2xy + 3xy = 3xy(x2 2y y2 2ay a2 + 1)

= 3xy[( x2 2x + 1) (y2 + 2ay + a2)]

= 3xy[(x 1)2 (y + a)2] = 3xy[(x 1) (y + a)][(x 1) + (y + a)] = 3xy( x 1 y a)(x 1 + y + a) b) Bi tp t luyn: Bi tp 1: Phn tch a thc x3 + 6x2 + 9x thnh nhn t. Bi tp 2: Phn tch a thc A = 2x2 - 3xy + y2 - x - 1 thnh nhn t: Bi tp 3: Phn tch a thc A = 8x4 - 2x3 - 3x2 - 2x - 1 thnh nhn t. 6. Phng php thm v bt mt cng mt hng t: Phng php: Phng php thm v bt cng mt hng t nhm s dng phng php nhm xut hin dng t nhn t chung hoc dng hng ng thc. i vi phng php ny mun nm chc th cch tt nht l lm tht nhiu bi tp, ch khng c dng tng qut. Lu : i vi ton phn tch a thc thnh nhn t "gim dn s m ca ly tha": x

3m+2 + x

3m+1 + 1 th u cha nhn t x2 + x + 1.

Do khi phn tch th phi ch lm xut hin dng x2 + x + 1 v cc du "+" c th thay bng du "-". a) Bi tp p dng: Bi tp 1: Phn tch a thc x4 + x2 + 1 thnh nhn t. Phn tch:

Tch x2 thnh 2x

2 x2 : (lm xut hin hng ng thc) Ta c x

4 + x

2 + 1 = x

4 + 2x

2 + 1 x2 = (x4 + 2x2 + 1) x2



Thm x v bt x: (lm xut hin hng ng thc v t nhn t chung) Ta c x

4 + x

2 + 1 = x

4 x + x2 + x + 1 = (x4 x) + (x2 + x + 1)

Gii x

4 + x

2 + 1 = x

4 x + x2 + x + 1

= (x4 x) + (x2 + x + 1)

= x(x 1)(x2 + x + 1) + (x2 + x + 1) = (x

2 + x + 1)(x

2 x + 1)

Bi tp 2: Phn tch a thc x5 + x4 + 1 thnh nhn t. Cch 1: Thm x

3 v bt x3 (lm xut hin hng ng thc v t nhn t chung)

Gii x

5 + x

4 + 1 = x

5 + x

4 + x

3 x3 + 1

= (x5 + x

4 + x

3 )+ (1 x3 )

= x3(x

2+ x + 1)+ (1 x )(x2+ x + 1)

= (x2+ x + 1)(x

3 x + 1 )

Cch 2: Thm x3, x

2, x v bt x3, x2, x (lm xut hin t nhn t chung)

Gii x

5 + x

4 + 1 = x

5 + x

4 + x

3 x3 + x2 x2 + x x + 1

= (x5 + x

4 + x

3) + ( x3 x2 x ) + (x2 + x + 1)

= x3(x

2 + x + 1) x(x2 + x + 1) + (x2 + x + 1)

= (x2 + x + 1)(x

3 x + 1 )

Ch : Cc a thc c dng x4 + x2 + 1, x5 + x + 1, x5 + x4 + 1, x7 + x5 + 1,.; tng qut nhng a thc dng x3m+2 + x3n+1 + 1 hoc x3 1, x6 1 u c cha nhn t x2 + x + 1. Bi tp 3: Phn tch a thc x4 + 4 thnh nhn t. Gi : Thm 2x2 v bt 2x2 : (lm xut hin hng ng thc)

Gii x

4 + 4 = x

4 + 4x

2 + 4 4x2 = (x2 + 2)2 (2x)2 = (x2 + 2 2x)( x2 + 2 + 2x)

Bi tp 4: Phn tch a thc x4 + 16 thnh nhn t. Gii Cch 1: x

4 + 4 = (x

4 + 4x

2 + 4) 4x2

= (x2 + 2)

2 (2x)2 = (x2 2x + 2)(x2 + 2x + 2)

Cch 2: x4 + 4 = (x

4 + 2x

3 + 2x

2) (2x3 + 4x2 + 4x) + (2x2 + 4x + 4)

= (x2 2x + 2)(x2 + 2x + 2)

Bi tp 5: Phn tich a thc x5 + x - 1 thnh nhn t Gii Cch 1: x5 + x - 1 = x5 - x4 + x3 + x4 - x3 + x2 - x2 + x - 1 = x

3(x

2 - x + 1) - x

2(x

2 - x + 1) - (x

2 - x + 1)

= (x2 - x + 1)(x

3 - x

2 - 1).

Cch 2: Thm va bt x2: x

5 + x - 1 = x

5 + x

2 - x

2 + x - 1 = x

2(x

3 + 1) - (x

2 - x + 1)

= (x2 - x + 1)[x

2(x + 1) - 1] = (x

2 - x + 1)(x

3 - x

2 - 1).

Bi tp 6: Phn tich a thc x7 + x + 1 thnh nhn t. Gii x

7 + x

2 + 1 = x

7 x + x2 + x + 1 = x(x6 1) + (x2 + x + 1)

= x(x3 1)(x3 + 1) + (x2+ x + 1)

= x(x3 + 1)(x - 1)(x

2 + x + 1) + ( x

2 + x + 1)

= (x2 + x + 1)(x

5 - x

4 x2 - x + 1)

Lu y: Cc a thc dng x3m + 1 + x3n + 2 + 1 nh x7 + x2 + 1, x4 + x5 + 1 u cha nhn t l x2 + x + 1.

Bi tp 7: Phn tch da thc 4x 4 +81 thnh nhn t.

Gii

Ta thm v bt vo a thc 4x 4 +81 hng t 36x 2 ta c:

4x4 +81 = 4x

4+36x

2 +81 -36x

2 = (2x

2+9)

2 (6x)2 = 2 22x 6x 9 2x 6x 9

Nhn xt: Trong trng hp ny dng cho a thc c hai hng t. Bi tp 8: Phn tch a thc x5 + x -1 thnh nhn t. Gii

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Ta thm bt x4, x3, x2 nh sau: x

5 + x - 1 = x

5 + x

4 +x

3 + x

2 - x

4 - x

3 - x

2 + x - 1

= (x5 - x

4 + x

3) + (x

4 - x

3 + x

2) (x2 - x + 1 )

= x3(x

2 - x + 1) + x

2(x

2 - x + 1) - (x

2 - x + 1)

= (x2 - x + 1)(x

3 + x

2 - 1)

Bi tp 9: Phn tch a thc sau thnh nhn t: A = a10 + a5 + 1 Gii A = (a

10 + a

9 + a

8) + (a

7 + a

6 + a

5) + (a

5 + a

4 + a

3) + (a

2 + a + 1) - (a

9 + a

8 + a

7) - (a

6 + a

5 + a

4) - (a

3

+ a2 + a)

= a8(a

2 + a + 1) + a

5(a

2 + a + 1) + a

3(a

2 + a + 1) - a

7(a

2 + a + 1) - a

4(a

2 + a + 1) - a(a

2 + a + 1)

b) Bi tp t luyn: Bi tp 1: Phn tch a thc A = x7 + x2 + 1 thnh nhn t. Bi tp 2: Phn tch a thc A = x5 + x4 + 1 thnh nhn t. Bi tp 3: Phn tch a thc A = x7 + x5 + 1 thnh nhn t. Bi tp 4: Phn tch a thc A = x8 + x7 + 1 thnh nhn t. Bi tp 5: Phn tch a thc A = x5 + x1 - 1 thnh nhn t. 7. Phng php i bin s (t n s ph): Phng php: Phng php ny thng dng a mt a thc bc cao v a thc c bc thp hn. Phng php ny khng c cng thc tng qut. Trong phng php ny, c trng hp c bit khi phn tch: Phn tch a thc i xng thnh nhn t. Lu : Cch gii mt s phng trnh. Cn s dng thm phng php thm bt hng t. Khi phn tch a thc i xng bc chn thnh nhn t th ta chia cho a thc cho x2 (hay l

t x2 lm nhn t chung), nhm hai hng t thch hp ri t n ph cho 1

xx

.

Cc a thc i xng bc l lun c nghim l - 1. Phn tch a thc thnh hai nhn t l (x + 1) v nhn t th 2 l a thc i xng bc chn. phn tch ht a thc th phn tch a thc th hai theo tng qut a thc i xng bc chn. Phng php ny dng n gin hn cc biu thc v a biu thc v dng gn hn.

a) Bi tp p dng: Bi tp 1: Phn tch a thc A = x4 + 4x3 + 5x2 + 4x + 1 thnh nhn t.

Gii (y l a thc i xng bc chn) Nhn thy x = 0 khng l nghim ca a thc trn. Ta t x2 lm nhn t chung. Khi :

2 2

2

1 1A = x x + + 4 x + + 5

x x

t: y = 1

xx

2

2 2

2

2 2

2

1 1y x x 2

x x

1y 2 x

x

Lc ny:

A = x2(y

2 + 4y + 3)

= x2(y + 3)(y + 1)

= 2 2 21 1

x x 3 x 1 x 3x 1 x x 1x x

Bi tp 2: Phn tch a thc x5 + 5x4 + 2x3 + 2x2 + 5x + 1 thnh nhn t. HD: y l a thc i xng bc l. Ta t nhn t (x + 1) v nhn t cn li l a thc bc chn th lm tng t.



Bi tp 3: Phn tch a thc sau thnh nhn t : x(x + 4)(x + 6)(x + 10) + 128

Gii Ta c: x(x + 4)(x + 6)(x + 10) + 128 = (x

2 + 10x)(x

2 + 10x + 24) + 128

t x2 + 10x + 12 = y. a thc a cho co dang: (y - 12)(y + 12) + 128 = y

2 - 16

= (y + 4)(y - 4) = (x2 + 10x + 16)(x

2 + 10x + 8)

= (x + 2)(x + 8)(x2 + 10x + 8)

Nhn xt: Nh phng php i bin ta a a thc bc 4 i vi x thnh a thc bc 2 i vi y. Bi tp 4: Phn tich a thc sau thanh nhn t : A = x4 + 6x3 + 7x2 - 6x + 1.

Gii Cch 1. Gi s x 0. Ta vit a thc di dang:

2 2

2

1 1A x x 6 x 7

x x

t 2 22

1 1x y x y 2

x x .

Do o: A = x

2(y

2 + 2 + 6y + 7) = x

2(y + 3)

2 = (xy + 3x)

2

= 2

221x x 3x x 3x 1

x

= (x

2 + 3x - 1)

2.

Dng phn tich nay cung ung vi x = 0. Cch 2. A = x4 + 6x3 - 2x2 + 9x2 - 6x + 1 = x4 + (6x3 -2x2) + (9x2 - 6x + 1) = x

4 + 2x

2(3x - 1) + (3x - 1)

2 = (x

2 + 3x - 1)

2.

b) Bi tp t luyn: Bi tp 1: Phn tch cc a thc sau thnh nhn t: x(x + 4)(x + 6)(x + 10) + 128 Bi tp 2: Phn tch cc a thc sau thnh nhn t: (x2 + 4x + 8)2 + 3x(x2 + 4x + 8) + 2x2 Bi tp 3: Phn tch cc a thc sau thnh nhn t: (x2 + x)2 + 4x2 + 4x - 12 Bi tp 4: Phn tch cc a thc sau thnh nhn t: x2 + 2xy + y2 + 2x + 2y - 15 Bi tp 5: Phn tch cc a thc sau thnh nhn t: (x2 + y2 + z2)(x + y + z)2 + (xy + yz + zx)2 8. Phng php xt gi tr ring: Phng php: Trc ht ta xc nh dng cc tha s cha bin ca a thc, ri gn cho cc bin cc gi tr c th xc nh tha s cn li. i vi cc bi ton dng ny th ta lun nhn bit s ging nhau v vi tr ca cc bin trong biu thc. Cch gii thng dng l s dng phng php l lun vi tr ca mt bin so vi cc bin cn li. a) Bi tp p dng: Bi tp 1: Phn tch a thc sau thnh nhn t: P = x2 (y - x) + y2(z - x) + z2(x - y) Gii Gi s thay x bi y th P = y2(y - z) + y2(z - y) = 0 Nh vy P cha tha s x - y. Ta li thy nu thay x bi y, thay y bi z, thay z bi x th P khng i (ta ni a thc P c th hon v vng quanh bi cc bin x, y, z. Do nu P cha tha s x - y th cng cha tha s y - z, z - x.

Vy P phi c dng: P = k(x - y)(y - z)(z - x). Ta thy k phi l hng s (khng cha bin) v P c bc 3 i vi tp hp cc bin x, y, z cn tch (x - y)(y - z)(z - x) cng c bc ba i vi tp hp cc bin x, y, z. V ng thc:

x2 (y - x) + y

2(z - x) + z

2(x - y) = k(x - y)(y - z)(z - x).

ng vi mi x, y, z nn ta gn cho cc bin x, y, z cc gi tr ring. Chng hn x = 2, y = 1, z = 0. Ta c k = - 1.

b) Bi tp t luyn:

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Bi tp 1: Phn tch a thc sau thnh nhn t: Q =a(b + c - a

2)

2 + b(c + a - b)

2 + c(a + b - c)

2 + (a + b - c)(b + c - a)(c + a - b)

Bi tp 2: Phn tch cc a thc sau thnh nhn t: M = a(m - a)

2 + b(m - b)

2 + c(m - c)

2 - abc, vi 2m = a + b + c.

Bi tp 3: Phn tch a thc sau thnh nhn t: A = (a + b + c)(ab + bc + ca) - abc

B = a(a + 2b)3 - b(2a + b)

3.

C = ab(a + b) - bc (b + c) + ac(a - c)

D = (a + b)(a2 - b

2) + (b + c)(b

2 - c

2) + (c + a)(c

2 - a

2)

E = a3(c - b

2) + b

3(a - c

2) + c

3(b - a

2) + abc(abc - 1)

F = a(b - c)3 + b(c - a)

3 + c(a - b)

3

G = a2b

2(a - b) + b

2c

2(b - c) + a

2c

2(c - a)

H = a4(b - c) + b

4(c - a) + c

4(a - b)

9. Phng php dng h s bt nh: Phng php: Gi s a thc f(x) = ax2 + bx + c c nghim m, n th a thc s c vit li:

f(x) = ax2 + bx + c = a(x - m)(x - n)

Sau ng nht h s c hai v ca phng trnh, tc l gii h phng trnh hoc nhm tm h s th ta s tm c cc h s m, n. a) Bi tp p dng: Bi tp 1: Phn tch a thc sau thnh nhn t: A = 3x2 + 4x + 1 Gii Gi s a v b l hai nghim ca a thc trn. Khi , ta vit li nh sau: 3x

2 + 4x + 1 = 3(x - a)(x - b)

3x2 + 4x + 1 = 3x2 + x(-3b - 3a) + 3ab ng nht h s ta c:

4a 1a b

3a 3b 4 31

1 3ab 1 bab 3

3

Vy A = 3x2 + 4x + 1 = 1

3 x 1 x x 1 3x 13

.

Bi tp 2: Phn tch a thc sau thnh nhn t: x4 + 6x3 + 7x2 + 6x + 1

Gii Nhn thy a thc trn khng c nghim hu t. Do ta s phn tch a thc trn thnh tch ca hai a thc bc hai. x

4 + 6x

3 + 7x

2 + 6x + 1 = (x

2 + ax + 1)(x

2 + cx + 1)

S dng phng php ng nht h s, ta c: x

4 + 6x

3 + 7x

2 + 6x + 1 = (x

2 + x + 1)(x

2 + 5x + 1)

Bi tp 3: Phn tch a thc sau thnh nhn t: x4 - 3x3 - x2 - 7x + 2. Gii

Nhn thy cc s 1, 2 u khng l nghim ca a thc, nn C khng c nghim hu t. Nh vy, nu a thc C phn tch c thnh nhn t th phi c dng. (x

2 + ax + b)(x

2 + cx + d) = x

4 + (a + c)x

3 + (ac + b + d)x

2 + (ad + bc)x + bd.

ng nht cc h s ca a thc ny vi a thc cho, ta c h phng trnh:

a c 3

ac b d 1

ad bc 7

bd 2

Gii h ny ta tm c (a; b; c; d) = (1; 2; -4; 1). Vy a thc cho c phn tch thnh: (x2 + x + 2)(x2 - 4x + 1). a thc ny khng phn tch thnh nhn t thm c na. Bi tp 4: Phn tich a thc sau thanh nhn t : x4 - 6x3 + 12x2 - 14x - 3



Gii Ta ln lt th cc nghim 1; 3 khng l nghim ca a thc , a thc khng co nghim nguyn cng khng co nghim hu t . Nh vy a thc trn p hn tich c thnh nhn t thi phi c dng: (x

2 + ax + b)(x

2 + cx + d) = x

4 +(a + c)x

3 + (ac+b+d)x

2 + (ad+bc)x + bd

= x4 - 6x

3 + 12x

2 - 14x + 3.

ng nht cc h s ta c:

Xt bd = 3 vi b, d Z, b { 1, 3}. Vi b = 3 th d = 1, h iu kin trn tr thanh

a c 6

ac 8

a 3c 14

2c = -14 - (-6) = -8.

Do o c = -4, a = -2. Vy x4 - 6x3 + 12x2 - 14x + 3 = (x2 - 2x + 3)(x2 - 4x + 1). b) Bi tp t luyn: Bi tp 1: Phn tch cc a thc sau thnh nhn t: 4x4 + 4x3 + 5x2 + 2x + 1 Bi tp 2: Phn tch cc a thc sau thnh nhn t: C = 3x

2 + 22xy + 11x + 37y + 7y

2 + 10

D = x4 - 7x

3 + 14x

2 - 7x + 1

E = x4 - 8x + 63

10. Phng php nhm nghim (thm): Phng php: nh l: Nu f(x) c nghim x = a th f(a) = 0. Khi , f(x) c mt nhn t l x - a v f(x) c th vit di dng f(x) = (x - a).q(x) Nh vy a thc f(x) s c nhn t l (x - a). Lc tch cc s hng ca f(x) thnh cc nhm, mi nhm u cha nhn t l x a. Lu : Phng php ny ch p dng nhiu cho a thc c h s nguyn:

P(x) = anxn + an-1x

n-1 + ... + a1x + a0.

H qu 1: Nu f(x) c tng cc h s bng 0 th f(x) c mt nghim l x = 1. T f(x) c mt nhn t l x 1. H qu 2: Nu f(x) c tng cc h s ca cc lu tha bc chn bng tng cc h s ca cc lu tha bc l th f(x) c mt nghim x = 1. T f(x) c mt nhn t l x + 1.

H qu 3: Nu f(x) c nghim nguyn x = a v f(1) v f(1) khc 0 th

f 1

a 1

v

f 1

a 1

u l s

nguyn.

Ngi ta chng minh nghim ca a thc: P(x) = anx

n + an-1x

n-1 + ... + a1x + a0.

L nghim ca hng t t do a0.

Ngi ta cng chng minh c nghim ca a thc c dng p

xq

, trong p l c ca a0 v

q l c ca hng t cao nht an. a) Bi tp p dng: Bi tp 1: Phn tch a thc f(x) = x3 + x2 + 4 thnh nhn t. Gii Ln lt kim tra vi x = 1, 2, 4, ta thy f(2) = (2)3 + (2)2 + 4 = 0. a thc f(x) c mt nghim x = 2, do n cha mt nhn t l x + 2. T , ta tch nh sau Cch 1 : f(x) = x

3 + 2x

2 x2 + 4 = (x3 + 2x2) (x2 4) = x2(x + 2) (x 2)(x + 2)

= (x + 2)(x2 x + 2).

Cch 2 : f(x) = (x3 + 8) + (x

2 4) = (x + 2)(x2 2x + 4) + (x 2)(x + 2)

= (x + 2)(x2 x + 2).

Cch 3 : f(x) = (x3 + 4x

2 + 4x) (3x2 + 6x) + (2x + 4)

= x(x + 2)2 3x(x + 2) + 2(x + 2) = (x + 2)(x2 x + 2).

Cch 4 : f(x) = (x3 x2 + 2x) + (2x2 2x + 4) = x(x2 x + 2) + 2(x2 x + 2)

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= (x + 2)(x2 x + 2).

Bi tp 2: Phn tch a thc f(x) = 4x3 - 13x2 + 9x - 18 thnh nhn t. Gii Cc c ca 18 l 1, 2, 3, 6, 9, 18. f(1) = 18, f(1) = 44, nn 1 khng phi l nghim ca f(x). D thy khng l s nguyn nn 3, 6, 9, 18 khng l nghim ca f(x). Ch cn 2 v 3. Kim tra ta thy 3 l nghim ca f(x). Do , ta tch cc hng t nh sau: f(x) = 4x

3 - 12x

2 - x

2 + 3x + 6x - 18

= 4x2(x - 3) - x(x - 3) + 6(x - 3)

=(x 3)(4x2 x + 6) Bi tp 3: Phn tch a thc f(x) = 3x3 - 7x2 + 17x - 5 thnh nhn t.

Gii Cc c ca 5 l 1, 5. Th trc tip ta thy cc s ny khng l nghim ca f(x). Nh vy

f(x) khng c nghim nghuyn. Xt cc s 1 5

; 3 3

, ta thy 1

3 l nghim ca a thc.

Do a thc c mt nhn t l 3x 1. Ta phn tch nh sau: f(x) = (3x3 x2) (6x2 2x) + (15x 5) = (3x 1)(x2 2x + 5). Bi tp 4: Phn tch a thc sau thnh nhn t: x3 5x2 + 8x 4. Gii Nhn thy a thc c 1 + (5) + 8 + (4) = 0 nn x = 1 l mt nghim ca a thc. a thc c mt nhn t l x 1. Ta phn tch nh sau : f(x) = (x

3 x2) (4x2 4x) + (4x 4) = x2(x 1) 4x(x 1) + 4(x 1)

= (x 1)( x 2)2 Bi tp 5: Phn tch a thc sau thnh nhn t: x3 5x2 + 3x + 9.

Gii Nhn thy, a thc x3 5x2 + 3x + 9 c 1 + 3 = 5 + 9 nn x = 1 l mt nghim ca a thc. a thc c mt nhn t l x + 1. Ta phn tch nh sau : f(x) = (x

3 + x

2) (6x2 + 6x) + (9x + 9) = x2(x + 1) 6x(x + 1) + 9(x + 1)

= (x + 1)( x 3)2



CHUYN 5 TP XC NH

1) Kin thc c bn: Bi ton: Cho biu thc: y = f(x), vi x l n s. nh ngha: Tp xc nh ca hm s l tp hp nhng gi tr lm cho biu thc c ngha. K hiu: D = {x| f(x) c ngha (iu kin)} 2) Tp xc nh ca mt s biu thc:

Biu thc:

f xA =

g x TX: D = {x| g(x) 0}

Biu thc: A = f x TX: D = {x| f(x) 0}

Ch : Nu nA = f x th Khi n l s l,vi mi x u tha mn. Khi n l s chn th f(x) 0.

Biu thc:

f xA =

g xc TX: D = {x| g(x) > 0}

Biu thc: A = f(x) c TX: D = R (vi f(x) = anxn + an-1x

n-1 + ... + a1x + a0)

3. Bi tp p dng: Bi tp 1: Tm tp xc nh ca biu thc sau:

y = 3x2 + 2x + 1

Gii iu kin xc nh D = R. Bi tp 2: Tm tp xc nh ca biu thc:

y = 3x2 + x - 1 +

1 - x

x + 2

Gii

iu kin xc nh: D = x | x + 2 0 = x | x -2 Bi tp 3: Tm tp xc nh ca biu thc:

y = x - 3 - x -1

Gii

iu kin xc nh: x - 3 0 x 3

x 3x -1 0 x 1

Bi tp 4: Tm iu kin xc nh ca biu thc:

1 3A = -

x - 2 x x + 2

Gii

iu kin xc nh:

x - 2 > 0 x > 2

x 0 x 0 x > 2

x + 2 > 0 x > -2

Bi tp 5: Tm tp xc nh ca biu thc:

a aT = a + ab

b b

Gii

iu kin xc nh: b 0 a < 0, b < 0

ab > 0 a > 0, b > 0

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4. Bi tp t luyn: Bi tp 1: Tm tp xc nh ca biu thc:

x 1 x - x x + xB = - -

2 2 x x +1 x -1


3x + 3P =

x + x + x +13 2


P = x - 2x +1 + x - 6x + 92 2


x + 2 x + 2 x + 1P = - +

x -1x - 2 x + 1


x -1 x +1 1P = - .

x - 2 x - 3x + 4 x + 4



CHUYN 6 RT GN BIU THC

1. Kin thc c bn: Dng khai trin ca mt s biu thc:

a - b = a + b a - b , vi a, b 0.

3 32 23 3 3a + b = a b a - ab + b+ 3 32 23 3 3a - b a - b a + ab + b= a a a , 3 vi a 0

a a , 2 vi a 0

Ch : Trc khi rt gn phi tm iu kin xc nh ca biu thc (nu c).

2 2A + A - B A - A - B

+ = A + B2 2

2

2

22

1 1 1= -

n n +1 n n +1

1= n +1 - n

n + n +1

k +111+ =

k + 2k k k + 2

1 1 1= -

n +1 n + n n +1 n n +1

1 1 1 11+ + = 1+ -

n n n +1n +1

2) Bi tp p dng: Bi tp 1: Rt gn biu thc:

A = (a + 3)(a - 3)(a2 + 6a + 9)( a

2 - 6a + 9)

Gii A = (a

2 - 9)(a + 3)

2(a - 3)

2 = (a

2 - 9)

3 = a

6 - 27a

4 + 243a

2 - 729

Lu : Bi ton ny c a v dng hng ng thc. Bi tp 2: Rt gn biu thc sau:

5 3 + 50 5 - 24 1 1A = - - +1

75 - 5 2 5 - 2 5 + 2

Gii

5 3 + 5 2 5 - 2 6 1 1A = - - +1

5 3 - 5 2 5 - 2 5 + 2

5 3 + 2 5 - 2 6 5 + 2 - 5 + 2 = - -1

35 3 - 2

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2

2 2

5 3 + 2 3 - 2 2 2 = - -1

35 3 - 2

5 3 - 2 3 - 22 2

= - -135 3 - 2

2 2 = -

3

Lu : Bi ton ny s dng phng php a tha s ra ngoi du cn v hng ng thc a2 - b

2.

Bi tp 3: Rt gn biu thc sau: 19 3 9 4

9 10 10

2 .27 +15.4 .9M =

6 .2 +12

( thi HSG min Bc nm 1997) Gii

18 919 9 18 8 19 9 18 9

9 9 10 10 10 19 9 20 10 18 9

2 .3 2 + 52 .3 + 5.3.2 .3 2 .3 + 5.2 .3 1M = = = =

2 .3 .2 + 4 .3 2 .3 + 2 .3 2 .3 .2 1+ 2.3 2

Lu : Bi ton ny s dng phng php a v dng chung ca ly tha t v mu. Bi tp 4: Rt gn biu thc sau:

6 44 2 3 2

8x - 27 y -1A = :

4x + 6x + 9 y + y + y +1

Gii

32 3 3 2

4 2 3 2

2 4 2 2

4 2

2x - 3 y -1 y + y + y +1A = :

4x + 6x + 9 y + y + y +1

2x - 3 4x + 6x + 9 2x - 3 = : y -1 =

4x + 6x + 9 y -1

Lu : Bi ton ny c a v dng hng ng thc. Bi tp 5: Rt gn biu thc sau:

3 31+ 1-

2 2A = +3 3

1+ 1+ 1- 1-2 2

Gii Ta c:

2

3 +13 4 + 2 31+ = =

2 4 4

2

3 -13 4 - 2 31- = =

2 4 4

Do , ta c:

2 2

3 +1 3 -1

2 2 3 +1 3 -1A = + = + = 1

2 3 2 33 +1 3 3 -1 3

2 2

Lu : Nhn bit bi ton ny l dng 2A = A xut hin mu.

Bi tp 6: Rt gn biu thc sau:



A = 4 + 15 10 - 6 4 - 15 Gii

2

A = 10 - 6 4 + 15 4 - 15 = 10 - 6 4 + 15

V 10 - 6 = 2 5 - 3 > 0 nn ta c:

2

A = 2 5 - 3 4 + 15 = 2 8 - 2 15 4 + 15 = 4 = 2

Lu : Bi ton ny c a v dng ng ca hng ng thc. Bi tp 7: Rt gn biu thc sau:

19 18 2 20A = 1978 1979 +1979 +...+1979 +1980 -1979 +1 Gii

19 18 2 20

20 20

20 20

A = 1979 -1 1979 +1979 + ... +1979 +1980 -1979 +1

= 1979 -1 -1979 +1

= 1979 -1+1979 +1 = 0

Lu : Bi ton ny c a v dng ng ca hng ng thc nng cao: an - bn. Bi tp 8: Rt gn biu thc sau:

2 x - 9 x + 3 2 x +1A = - -

x - 5 x + 6 x - 2 3 - x

Gii iu kin xc nh: x 0, x 4; x 9. Ta c:

Mu thc chung l: x - 5 x + 6 = x - 2 x - 3

2 x - 9 - x - 9 + 2 x -1 x - 2 x - x - 2A = =

x - 2 x - 3 x - 2 x - 3

x +1 x - 2 x +1 = =

x - 3x - 3 x - 2

Lu : Nhn bit dng bi ton ny l phi quy ng tm mu thc chung. Sau tm nhn t chung ca t v mu. Bi tp 9: Rt gn biu thc sau:

1 1 1 1A = + : -

x +1 x -1 x -1 x +1

Gii iu kin: x > 1.

2 2

2

2

x +1 + x -1 x +1 - x -1A = :

x -1 x -1

x +1 + x -1x +1 + x -1 = =

x +1- x +1x +1 - x -1

= x + x -1

Lu : Bi ton ny s dng cch quy ng v rt gn nhn t chung (ging nhau) ca t v mu. Bi tp 10: Rt gn biu thc sau:

2

3 2

2y + 5y + 2A =

2y + 9y +12y + 4

( thi HSG ton quc nm 1978)

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Gii Ta c t thc: 2y

2 + 5y + 2 = (2y

2 + 4y) + (y + 2) = 2y(y + 2) + (y + 2)

= (y + 2)(2y + 1)

Ta c mu thc: 2y

3 + 9y

2 + 12y + 4 = (2y

3 + 4y

2) + (5y

2 + 10y) + (2y + 4)

= 2y2(y + 2) + 5y(y + 2) + 2(y + 2)

= (y + 2)(2y2 + 5y + 2) = (y + 2)

2(2y + 1)

V 2y2 + 5y + 2 = (y + 2)(2y + 1)

Do iu kin bi ton l: 1

y -2; y -2

Suy ra:

2

y + 2 2y +1 1A = =

y + 2y + 2 2y +1

Lu : Bi ton ny khng th a hng ng thc m cn phi phn tch t v mu thc xut hin nhn t chung. Bi tp 11: Rt gn biu thc sau:

A = 2 + 3. 2 + 2 + 3 . 2 + 2 + 2 + 3 . 2 - 2 + 2 + 3

Gii Ta c:

2 + 2 + 2 + 3 . 2 - 2 + 2 + 3 = 4 - 2 + 2 + 3 = 2 - 2 + 3

2 + 2 + 3 . 2 - 2 + 3 = 4 - 2 + 3 = 2 - 3

A = 2 + 3. 2 - 3 = 1

Lu : Bi ton ny c hiu l "cn chng cn" th ta phi gii quyt 2A = A t trong ra

ngoi gim dn bc ca cn thc. Bi tp 12: Rt gn biu thc sau:

1 x xA = + :

x x +1 x + x

Gii iu kin: x > 0.

x +1+ x x x +1+ xA = : =

xx x +1 x x +1

Lu : Bi ton ny nhn bit c ngy l phi quy ng v tm nhn t chung ca t v mu. Bi tp 13: Rt gn biu thc sau:

4 - 2 3A =

6 - 2

Gii

2

3 -14 - 2 3 3 -1 2A = = =

26 - 2 2 3 -1 2 3 -1

Lu : Bi ton ny c a v dng hng ng thc v 2A = A sao cho xut hin nhn t

ging nhau ca t v mu. Bi tp 14: Rt gn biu thc sau:

A = 3 2 + 6 6 - 3 3 Gii



2

3 4 - 2 3A = 6 3 +1 3 2 - 3 = 6 3 +1

2

3 3 -1 6 3 +1 3 3 -1 = 6 3 +1 =

2 2

= 3 3 +1 3 -1 = 6

Lu : Nhn bit c dng ton 2A = A v sau a v dng hng ng thc.


a + b - 2 ab 1A = :

a - b a + b

Gii iu kin: a > 0, b > 0, a b.

2 2 2

a - 2 a. b + b a - b1 1A = : = :

a - b a + b a - b a + b

= a - b a + b = a - b


1x

1x

1xx

1x

1xx

2x:1P vi 1x0

Gii

1x

1x

1xx

1x

1xx

2x:1P

)1x)(1x(

1x

1xx

1x

)1xx)(1x(

2x:1P

1x

1

1xx

1x

)1xx)(1x(

2x:1P

)1xx)(1x(

1xx

)1xx)(1x(

)1x)(1x(

)1xx)(1x(

2x:1P

)1xx)(1x(

)1xx()1x(2x:1P

)1xx)(1x(

xx:1P

)1xx)(1x(

)1x.(x:1P

1xx

x:1P

Suy ra: x

1xxP

Lu : Ta nhn bit ngay l s dng hng ng thc a2 - b2 v (a - b)2.

3) Bi tp t luyn Bi tp1: Cho biu thc:

a + 2 5P = - +

a + 3 a + a - 6

1

2 - a

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a) Rt gn P b) Tm gi tr ca a P < 1 Bi tp 2: Cho biu thc:

P = x x + 3 x + 2 x + 2

1- : + +x +1 x - 2 3 - x x - 5 x + 6

a) Rt gn P b)Tm gi tr ca x P < 0. Bi tp 3: Cho biu thc:

P = x -1 1 8 x 3 x - 2

- + : 1-9x -13 x -1 3 x +1 3 x +1

a) Rt gn P

b) Tm cc gi tr ca x P = 5

6

Bi tp 4: Cho biu thc:

P = a 1 2 a

1+ : -a +1 a -1 a a + a - a -1

a) Rt gn P b) Tm gi tr ca a P < 1

c) Tm gi tr ca P nu a = 19 -8 3


P=2 3 3a (1- a) 1- a 1+ a

: + a . - a1+ a 1- a 1+ a

a) Rt gn P

b) Xt du ca biu thc M = a.1

P -2


P = x +1 2x + x x +1 2x + x

+ -1 : 1+ -2x +1 2x -1 2x +1 2x -1

a) Rt gn P

b) Tnh gi tr ca P khi x 1= 3 + 2 22


P = 2 x 1 x

- : 1+x +1x x + x - x -1 x -1

a) Rt gn P b) Tm x P0 Bi tp 8: Cho biu thc:

P=3

3

2a +1 a 1+ a- . - a

a + a +1 1+ aa

a) Rt gn P

b) Xt du ca biu thc P 1- a


P = 1- a a 1+ a a

+ a . - a1- a 1+ a

a) Rt gn P

b) Tm a P < 7 - 4 3




P = 2 x x 3x + 3 2 x - 2

+ - : -1x - 9x + 3 x - 3 x - 3

a) Rt gn P

b) Tm x P < 1

2

c) Tm gi tr nh nht ca P Bi tp 11: Cho biu thc:

P = x - 3 x 9 - x x - 3 x - 2

-1 : - -x - 9 x + x - 6 2 - x x + 3

a) Rt gn P b) Tm gi tr ca x P < 1 Bi tp 12: Cho biu thc:

P = 15 x -11 3 x - 2 2 x + 3

+ -x + 2 x - 3 1- x x + 3

a) Rt gn P

b) Tm cc gi tr ca x P = 1

2

c) Chng minh P 2

3


P=2

2

2 x x m+ -

4x - 4mx + m x - m, (vi m > 0)

a) Rt gn P b) Tnh x theo m P = 0. c) Xc nh cc gi tr ca m gi tr x tm c cu b tho mn iu kin x > 1. Bi tp 14: Cho biu thc:

P =2a + a 2a + a

- +1a - a +1 a

a) Rt gn P b) Bit a >1 Hy so snh P vi |P|. c) Tm a P = 2 d) Tm gi tr nh nht ca P. Bi tp 15: Cho biu thc:

P = a +1 ab + a a +1 ab + a

+ -1 : - +1ab +1 ab -1 ab +1 ab -1

a) Rt gn P

b) Tnh gi tr ca P nu a = 2 - 3 v b = 3 -1

1+ 3

c) Tm gi tr nh nht ca P nu a + b = 4


P = a a -1 a a +1 1 a +1 a -1

- + a - +a - a a + a a a -1 a +1

a) Rt gn P b) Vi gi tr no ca a th P = 7 c) Vi gi tr no ca a th P > 6 Bi tp 17: Cho biu thc:

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P =

2

a - b + 4 ab a b - b a.

a + b ab

a) Tm iu kin P c ngha. b) Rt gn P

c) Tnh gi tr ca P khi a = 2 3 v b = 3


P = 2a + a -1 2a a - a + a a - a

1+ -1- a 1- a a 2 a -1

a) Rt gn P

b) Cho P =6

1+ 6 tm gi tr ca a

c) Chng minh rng P > 2

3


P = a -1 . a - b3 a 3a 1

- + :a + ab + b a a - b b a - b 2a + 2 ab + 2b

a) Rt gn P b) Tm nhng gi tr nguyn ca a P c gi tr nguyn

Bi tp 20:

1a

1

1a

1:

1a

aa

)1a)(2a(

2a3aP ( vi 1a0 )

a) Rt gn P.

b) Tnh gi tr ca P khi 324a .

c) Tm a sao cho 1P

1 .



CHUYN 7 CC PHNG PHP BIN I BIU THC

1. Bin i biu thc nguyn: a. Kin thc c bn: Phng php: Ta bin i t mt v ca ng thc (p dng hng ng thc) chuyn thnh biu thc bng v cn li. Xt tnh cht ca mt s biu thc c bit a ra cch phn tch ng theo yu cu bi ton. p dng bin i theo mt v tr thnh v cn li hoc hai v cng v mt kt qu c th. b) Bi tp p dng: Bi tp 1: Chng minh cc ng thc sau: a) a

3 + b

3 + c

3 - 3abc = (a + b+ c)(a

2 + b

2 + c

2 - ab - bc - ca)

b) (a + b + c)3 - a

3 - b

3 - c

3 = 3(a + b)(b + c)(c + a)

Gii a) a

3 + b

3 + c

3 - 3abc = (a + b)

3 + c

3 - 3abc - 3a

2b - 3ab

2

= (a + b+ c)[(a + b)2 - (a + b)c + c

2] - 3ab(a + b + c)

= (a + b+ c)[(a + b)2 - (a + b)c + c

2 - 3ab]

= (a + b+ c)(a2 + b

2 + c

2 - ab - bc - ca). (pcm)

b) (a + b + c)3 - a

3 - b

3 - c

3 = [(a + b + c)

3 - a

3] - (b

3 + c

3)

= (b + c)([(a + b + c)2 + (a + b + c)a + a

2] - (b + c)(b

2 - bc + c

2)

= (b + c)(3a2 + 3ab + 3bc + 3ca)

= 3(b + c)[(a(a + b) + c(a + b)]

= 3(a + b)(b + c)(c + a). (pcm) Bi tp 2: Cho x + y + z = 0. Chng minh rng: 10(x7 + y7 + z7) = 7(x2 + y2 + z2)(x5 + y5 + z5)

Gii T x + y + z = 0. Suy ra: z = x(x + y). Ta c:

x5 + y

5 + z

5 = x

5 + y

5 - (x + y)

5

= -5(x4y + 2x

3y

2 + xy

4)

= -5xy(x3 + 2x

2y + 2xy

2 + y

3)

= -5xy[(x + y)(x2 + y

2 - xy) + 2xy(x + y)]

= -5xy(x + y)(x2 + y

2 + xy) (1)

v

x2 + y

2 + z

2 = x

2 + y

2 + (x + y)

2

= 2(x2 + y

2 + xy) (2)

m

x7 + y

7 + z

7 = x

7 + y

7 - (x + y)

7

= -7(x6y + 3x

5y

2 + 5x

4y

3 + 5x

3y

4 + 3x

2y

5 + xy

6)

= -7xy(x + y)(x2 + y

2 + xy)

(3)

T (1), (2) v (3) ta c iu phi chng minh. c) Bi tp t luyn: Bi tp 1: Cho a + b + c = 0. Chng minh rng: a) a

3 + b

3 + c

3 = 3abc.

b) 2(a4 + b

4 + c

4) = (a

2 + b

2 + c

2)

2.

Bi tp 2: Chng minh cc hng ng thc sau: a) (a

2 + b

2)(c

2 + d

2) = (ac + bd)

2 + (ad - bc)

2

b) a3 + b

3 + c

3 - 3abc = (a + b + c)(a

2 + b

2 + c

2 - ab - bc - ac)

Bi tp 3: Chng minh rng nu cc s a, b, c tha mn: a

4 + b

4 + (a - b)

4 = c

4 + d

4 + (c - d)

4 th a

2 + b

2 + (a - b)

2 = c

2 + d

2 + (c - d)

2

Bi tp 4: Cho a3 - 3ab2 = 19, b3 - 3a2b = 98. Tnh P = a2 + b2. Bi tp 5: Cho a, b l hai s tha mn iu kin: a2 - 3ab + 2b2 + a - b = a2 - 2ab + b2 - 5a + 7b = 0.

Chng minh rng: ab - 12a + 15b = 0.

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Bi tp 6: Khai trin biu thc: a4 + (a + b)4 thnh dng 2K + 1 v phn tch K thnh tch cc tha s. Bi tp 7:

a) Chng minh ng thc: x + y + |x - y| = max{x, y}, vi x, y R.

b) Chng minh ng thc: a + b a - b 2 a + b a - b 2 1 1 1

+ - + + + = 4max , , ab ab c ab ab c a b c

vi a, b, c 0. Bi tp 8: Tm tt c cc s nguyn dng n sao cho: a thc x3n+1 + x2n + 1 chia ht cho a thc x2 + x + 1. Bi tp 9: Cho x + y + z = 0. Chng minh rng: 2(x5 + y5 + z5) = 5xyz(x2 + y2 + z2).

( thi HSG tnh nm hc 2005 - 2006) Bi tp 10: Cho a2 - b2 = 4c2. Chng minh rng: (5a - 3b + 8c)(5a - 3b - 8c) = (3a - 5b)2.

2. Bin i hu t a. Kin thc c bn: Phng php: S dng cc bin i thng thng a n kt lun theo yu cu bi ton. Vn dng gi thit chn hng gii nhanh v chnh xc. i vi cc bi ton c s m bc n ca cc hng t, c th s dng phng php quy np. Lu : y l kin thc cn thit chng minh cc bt ng thc hay tm GTLN, GTNN. b) Bi tp p dng: Bi tp 1: Cho a + b + c = 0; a, b, c 0. Chng minh ng thc:

2 2 2

1 1 1 1 1 1+ + = + +

a b c a b c

Gii Ta c:

2

2 2 2

1 1 1 1 1 1 1 1 1+ + = + + + 2 + +

a b c a b c ab bc ca

2 2 2

2 2 2

2 a + b + c1 1 1= + + +

a b c abc

1 1 1= + + .

a b c

Suy ra iu phi chng minh. Bi tp 2: Chng minh rng nu a, b, c khc nhau th:

b -c c -a a - b 2 2 2

+ + = + +a - b a -c b -c b -a c - a c - b a - b b -c c -a

Gii Bin i v tri:

a - c - a - b b - a - b - c c - b - c - ab - c c - a a - b+ + = + +

a - b a - c b - c b - a c - a c - b a - b a - c b - c b - a c - a c - b

1 1 1 1 1 1 = - + - + -

a - b a - c b - c b - a c - a c - b

2 2 2

= + +a - b b - c c - a

c) Bi tp t luyn:

Bi tp 1: Cho xyz = 1. Chng minh rng: 1 1 1

+ + =11+ x + xy 1+ y + yz 1+ z + zx

Bi tp 2: Cho a, b, c, d l cc s thc tha mn iu kin:

a c a + c= = , a.c 0.

b d 3b -d Chng minh rng: b2 = d2.

Bi tp 3: Cho a, b, c l cc s thc tha mn iu kin 1 1 1

a + = b + = c +b c a



a) Cho a = 1. Tnh b, c.

b) Chng minh rng nu a, b, c i mt khc nhau th a2b2c2 = 1. c) Chng minh rng nu a, b, c dng th a = b = c.

Bi tp 4:

a) Cho a, b, c tha abc 0 v ab + bc + ca = 0. Tnh a + b b + c c + a

P =abc

.

b) Cho a, b, c tha (a + b)(b + c)(c + a) 0 v 2 2 2 2 2 2a b c a b c

+ + = + +a + b b + c c + a b + c c + a a + b

.

Chng minh rng: a = b = c.

Bi tp 5: Cho x > 0 tha iu kin: 22

1x + = 7

x. Tnh gi tr ca biu thc 5

5

1P = x +

x.

Bi tp 6: Cho a, b, x, y l cc s thc tha mn x2 + y2 = 1 v 4 4x y 1

+ =a b a + b

.

Chng minh rng:

2006 2006

10031003 1003

x y 1+ =

a b a + b.

Bi tp 7: Cho a, b, c l cc s thc tha mn b c v 2 2ac - b bd - c

= = k.a - 2b + c b - 2c + d

Chng minh: ad - bc

k =a - b - c + d

.

Bi tp 8: Cho a, b, c l cc s thc i mt khc nhau v khc 0, tha mn iu kin a + b + c = 0.

Chng minh rng: a b c b -c c -a a - b

+ + + + = 9.b -c c - a a - b a b c

Bi tp 9: n gin biu thc:

3 4 53 3 2 2

1 1 1 1 1 1+ +

a + b a + b a + ba + b a + b a + b

Bi tp 10: Chng minh rng nu: (a2 + b2)(x2 + y2) = (ax + by)2 v x, y khc 0 th a b

=x y

.

Bi tp 11: Chng minh rng nu: (a2 + b2 + c2)(x2 + y2 + z2) = (ax + by + cz)2 v x, y, z khc 0

th a b c

= =x y z

.

Bi tp 12: Cho a, b, khc 0 tha mn a + b = 1. Chng minh rng:

3 3 2 2

2 ab - 2a b+ =

b -1 a -1 a b +3

3. Bin i v t

a. Kin thc c bn: Phng php: p dng hng ng thc bin i. Mt s bng lin quan n cn bc n, c th s dng phng php quy np a n kt lun theo yu cu bi ton. Ging cc bi ton rt gn.

b) Bi tp p dng:

Bi tp 1: Chng minh ng thc:

4 449 20 60 49 20 60

32

Gii

2 24 44 4 5 + 2 6 + 5- 2 649 + 20 60 + 49 - 20 60

VT 1 = =2 2

4 44 43 + 2 + 3 - 2

=2

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3 + 2 + 3 - 2

= = 2 32

Bi tp 2: Cho a 0. Chng minh rng: 2 2 2

2 2

a - a a + a- + a +1 = a -1

a + a +1 a - a +1

Gii

Hng dn: t n ph: a = x 0

c) Bi tp t luyn:

Bi tp 1: Cho a, b > 0 v c 0. Chng minh rng: 1 1 1

0 a b a c b ca b c

Bi tp 2: Chng minh rng cc s sau l s nguyn:

a) 2 3+ 5- 3+ 48

A =6 + 2

b) 3 348 48

B 1 19 9

Bi tp 3: Cho x l s nguyn. Chng minh rng nu x khng phi l s chnh phng th x l s v t.

Bi tp 4: Chng minh rng s = 2 + 2 + 3 - 6 -3 2 + 3 l mt nghim ca phng

trnh x4 - 16x

2 + 32 = 0.

Bi tp 5: Chng minh rng nu

3 3 3ax by cz

1 1 11

x y z

th 2 2 2 3 3 33 ax by cz a b c .

Bi tp 6: Chng minh rng nu 3 3 3 3a b c a b c th vi s nguyn dng l n ta c n n n na b c a b c .

Bi tp 7: Cho 2 2xy 1 x 1 y a Tnh 2 2S x 1 y y 1 x .

Bi tp 8: Cho a > 0 v 24 2 2 0.a a Chng minh rng: 4 2

a 12

a a 1 a



CHUYN 7 CC PHNG PHP BIN I BIU THC

1. Bin i biu thc nguyn: a. Kin thc c bn: Phng php: Ta bin i t mt v ca ng thc (p dng hng ng thc) chuyn thnh biu thc bng v cn li. Xt tnh cht ca mt s biu thc c bit a ra cch phn tch ng theo yu cu bi ton. p dng bin i theo mt v tr thnh v cn li hoc hai v cng v mt kt qu c th. b) Bi tp p dng: Bi tp 1: Chng minh cc ng thc sau: a) a

3 + b

3 + c

3 - 3abc = (a + b+ c)(a

2 + b

2 + c

2 - ab - bc - ca)

b) (a + b + c)3 - a

3 - b

3 - c

3 = 3(a + b)(b + c)(c + a)

Gii a) a

3 + b

3 + c

3 - 3abc = (a + b)

3 + c

3 - 3abc - 3a

2b - 3ab

2

= (a + b+ c)[(a + b)2 - (a + b)c + c

2] - 3ab(a + b + c)

= (a + b+ c)[(a + b)2 - (a + b)c + c

2 - 3ab]

= (a + b+ c)(a2 + b

2 + c

2 - ab - bc - ca). (pcm)

b) (a + b + c)3 - a

3 - b

3 - c

3 = [(a + b + c)

3 - a

3] - (b

3 + c

3)

= (b + c)([(a + b + c)2 + (a + b + c)a + a

2] - (b + c)(b

2 - bc + c

2)

= (b + c)(3a2 + 3ab + 3bc + 3ca)

= 3(b + c)[(a(a + b) + c(a + b)]

= 3(a + b)(b + c)(c + a). (pcm) Bi tp 2: Cho x + y + z = 0. Chng minh rng: 10(x7 + y7 + z7) = 7(x2 + y2 + z2)(x5 + y5 + z5)

Gii T x + y + z = 0. Suy ra: z = x(x + y). Ta c:

x5 + y

5 + z

5 = x

5 + y

5 - (x + y)

5

= -5(x4y + 2x

3y

2 + xy

4)

= -5xy(x3 + 2x

2y + 2xy

2 + y

3)

= -5xy[(x + y)(x2 + y

2 - xy) + 2xy(x + y)]

= -5xy(x + y)(x2 + y

2 + xy) (1)

v

x2 + y

2 + z

2 = x

2 + y

2 + (x + y)

2

= 2(x2 + y

2 + xy) (2)

m

x7 + y

7 + z

7 = x

7 + y

7 - (x + y)

7

= -7(x6y + 3x

5y

2 + 5x

4y

3 + 5x

3y

4 + 3x

2y

5 + xy

6)

= -7xy(x + y)(x2 + y

2 + xy)

(3)

T (1), (2) v (3) ta c iu phi chng minh. c) Bi tp t luyn: Bi tp 1: Cho a + b + c = 0. Chng minh rng: a) a

3 + b

3 + c

3 = 3abc.

b) 2(a4 + b

4 + c

4) = (a

2 + b

2 + c

2)

2.

Bi tp 2: Chng minh cc hng ng thc sau: a) (a

2 + b

2)(c

2 + d

2) = (ac + bd)

2 + (ad - bc)

2

b) a3 + b

3 + c

3 - 3abc = (a + b + c)(a

2 + b

2 + c

2 - ab - bc - ac)

Bi tp 3: Chng minh rng nu cc s a, b, c tha mn: a

4 + b

4 + (a - b)

4 = c

4 + d

4 + (c - d)

4 th a

2 + b

2 + (a - b)

2 = c

2 + d

2 + (c - d)

2

Bi tp 4: Cho a3 - 3ab2 = 19, b3 - 3a2b = 98. Tnh P = a2 + b2. Bi tp 5: Cho a, b l hai s tha mn iu kin: a2 - 3ab + 2b2 + a - b = a2 - 2ab + b2 - 5a + 7b = 0.

Chng minh rng: ab - 12a + 15b = 0.

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Bi tp 6: Khai trin biu thc: a4 + (a + b)4 thnh dng 2K + 1 v phn tch K thnh tch cc tha s. Bi tp 7:

a) Chng minh ng thc: x + y + |x - y| = max{x, y}, vi x, y R.

b) Chng minh ng thc: a + b a - b 2 a + b a - b 2 1 1 1

+ - + + + = 4max , , ab ab c ab ab c a b c

vi a, b, c 0. Bi tp 8: Tm tt c cc s nguyn dng n sao cho: a thc x3n+1 + x2n + 1 chia ht cho a thc x2 + x + 1. Bi tp 9: Cho x + y + z = 0. Chng minh rng: 2(x5 + y5 + z5) = 5xyz(x2 + y2 + z2).

( thi HSG tnh nm hc 2005 - 2006) Bi tp 10: Cho a2 - b2 = 4c2. Chng minh rng: (5a - 3b + 8c)(5a - 3b - 8c) = (3a - 5b)2.

2. Bin i hu t a. Kin thc c bn: Phng php: S dng cc bin i thng thng a n kt lun theo yu cu bi ton. Vn dng gi thit chn hng gii nhanh v chnh xc. i vi cc bi ton c s m bc n ca cc hng t, c th s dng phng php quy np. Lu : y l kin thc cn thit chng minh cc bt ng thc hay tm GTLN, GTNN. b) Bi tp p dng: Bi tp 1: Cho a + b + c = 0; a, b, c 0. Chng minh ng thc:

2 2 2

1 1 1 1 1 1+ + = + +

a b c a b c

Gii Ta c:

2

2 2 2

1 1 1 1 1 1 1 1 1+ + = + + + 2 + +

a b c a b c ab bc ca

2 2 2

2 2 2

2 a + b + c1 1 1= + + +

a b c abc

1 1 1= + + .

a b c

Suy ra iu phi chng minh. Bi tp 2: Chng minh rng nu a, b, c khc nhau th:

b -c c -a a - b 2 2 2

+ + = + +a - b a -c b -c b -a c - a c - b a - b b -c c -a

Gii Bin i v tri:

a - c - a - b b - a - b - c c - b - c - ab - c c - a a - b+ + = + +

a - b a - c b - c b - a c - a c - b a - b a - c b - c b - a c - a c - b

1 1 1 1 1 1 = - + - + -

a - b a - c b - c b - a c - a c - b

2 2 2

= + +a - b b - c c - a

c) Bi tp t luyn:

Bi tp 1: Cho xyz = 1. Chng minh rng: 1 1 1

+ + =11+ x + xy 1+ y + yz 1+ z + zx

Bi tp 2: Cho a, b, c, d l cc s thc tha mn iu kin:

a c a + c= = , a.c 0.

b d 3b -d Chng minh rng: b2 = d2.

Bi tp 3: Cho a, b, c l cc s thc tha mn iu kin 1 1 1

a + = b + = c +b c a



a) Cho a = 1. Tnh b, c.

b) Chng minh rng nu a, b, c i mt khc nhau th a2b2c2 = 1. c) Chng minh rng nu a, b, c dng th a = b = c.

Bi tp 4:

a) Cho a, b, c tha abc 0 v ab + bc + ca = 0. Tnh a + b b + c c + a

P =abc

.

b) Cho a, b, c tha (a + b)(b + c)(c + a) 0 v 2 2 2 2 2 2a b c a b c

+ + = + +a + b b + c c + a b + c c + a a + b

.

Chng minh rng: a = b = c.

Bi tp 5: Cho x > 0 tha iu kin: 22

1x + = 7

x. Tnh gi tr ca biu thc 5

5

1P = x +

x.

Bi tp 6: Cho a, b, x, y l cc s thc tha mn x2 + y2 = 1 v 4 4x y 1

+ =a b a + b

.

Chng minh rng:

2006 2006

10031003 1003

x y 1+ =

a b a + b.

Bi tp 7: Cho a, b, c l cc s thc tha mn b c v 2 2ac - b bd - c

= = k.a - 2b + c b - 2c + d

Chng minh: ad - bc

k =a - b - c + d

.

Bi tp 8: Cho a, b, c l cc s thc i mt khc nhau v khc 0, tha mn iu kin a + b + c = 0.

Chng minh rng: a b c b -c c -a a - b

+ + + + = 9.b -c c - a a - b a b c

Bi tp 9: n gin biu thc:

3 4 53 3 2 2

1 1 1 1 1 1+ +

a + b a + b a + ba + b a + b a + b

Bi tp 10: Chng minh rng nu: (a2 + b2)(x2 + y2) = (ax + by)2 v x, y khc 0 th a b

=x y

.

Bi tp 11: Chng minh rng nu: (a2 + b2 + c2)(x2 + y2 + z2) = (ax + by + cz)2 v x, y, z khc 0

th a b c

= =x y z

.

Bi tp 12: Cho a, b, khc 0 tha mn a + b = 1. Chng minh rng:

3 3 2 2

2 ab - 2a b+ =

b -1 a -1 a b +3

3. Bin i v t

a. Kin thc c bn: Phng php: p dng hng ng thc bin i. Mt s bng lin quan n cn bc n, c th s dng phng php quy np a n kt lun theo yu cu bi ton. Ging cc bi ton rt gn.

b) Bi tp p dng:

Bi tp 1: Chng minh ng thc:

4 449 20 60 49 20 60

32

Gii

2 24 44 4 5 + 2 6 + 5- 2 649 + 20 60 + 49 - 20 60

VT 1 = =2 2

4 44 43 + 2 + 3 - 2

=2

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3 + 2 + 3 - 2

= = 2 32

Bi tp 2: Cho a 0. Chng minh rng: 2 2 2

2 2

a - a a + a- + a +1 = a -1

a + a +1 a - a +1

Gii

Hng dn: t n ph: a = x 0

c) Bi tp t luyn:

Bi tp 1: Cho a, b > 0 v c 0. Chng minh rng: 1 1 1

0 a b a c b ca b c

Bi tp 2: Chng minh rng cc s sau l s nguyn:

a) 2 3+ 5- 3+ 48

A =6 + 2

b) 3 348 48

B 1 19 9

Bi tp 3: Cho x l s nguyn. Chng minh rng nu x khng phi l s chnh phng th x l s v t.

Bi tp 4: Chng minh rng s = 2 + 2 + 3 - 6 -3 2 + 3 l mt nghim ca phng

trnh x4 - 16x

2 + 32 = 0.

Bi tp 5: Chng minh rng nu

3 3 3ax by cz

1 1 11

x y z

th 2 2 2 3 3 33 ax by cz a b c .

Bi tp 6: Chng minh rng nu 3 3 3 3a b c a b c th vi s nguyn dng l n ta c n n n na b c a b c .

Bi tp 7: Cho 2 2xy 1 x 1 y a Tnh 2 2S x 1 y y 1 x .

Bi tp 8: Cho a > 0 v 24 2 2 0.a a Chng minh rng: 4 2

a 12

a a 1 a



CHUYN 9 TNH GI TR CA BIU THC

1. Kin thc c bn: Phng php: Thng thng gii dng ton ny ta phi rt gn trc. Sau thay cc gi tr ca bin vo v tnh ton.

p dng mt s tnh cht ca cc biu thc, phn thc, hng ng thc phn tch bi ton. 2. Bi tp p dng:

Bi tp 1: Tnh gi tr ca biu thc: M = 7260136

Gii Ta c:

M = 2

2136 7260 11 2.11. 15 15 11 15

Vy: M = 11 + 15

Bi tp 2: Tnh gi tr ca biu thc: M = 15281528

Gii Ta c:

323535

55.3.2355.3.2315281528M2222

Vy: M = - 2 3

Bi tp 3: Tnh gi tr ca biu thc: N = 5122935

Gii Ta c:

115555215

526535355122935N

2

2

Vy: N = 1

Bi tp 4: Tnh gi tr ca biu thc: Q = xx21x2xx21x2 22 ti x = 4:

Gii Ta c:

x21xx1xx x)1x(x21xx)1x(x21x

xx21x2xx21x2Q 22

Thay x = 4, ta c: Q = 4.

Bi tp 5: Tnh gi tr ca biu thc: M = 48 - 752 + 108 - 1477

1

Gii Ta c:

M = 48 - 752 + 108 - 1477

1 = 3.16 - 3.252 + 36.3 - 3.49

7

1

= 34 - 310 + 36 - 3 = - 3

Bi tp 6: Tnh gi tr ca biu thc: S =

x1x

)x3(x1x3xx3 2

ti x = 10.

Gii Ta c:

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0xxx

1x

1xxx

1x

xx x

1x

xx

x1x

xx.x3x.x3xx

1x

)x3(x1x3xxS

33

3 3223

3 2

Vy: S = 0

Bi tp 7: Tnh gi tr ca biu thc: M = 1x21x 84 (vi x > 1), ti x = 256.

Gii Ta c:

M = 882884 x11x11x1x21x Vy: M = 2.

c) Bi tp t luyn: Bi tp 1: Cho cc s thc dng a v b tho mn: a100 + b100 = a101 + b101 = a102 + b102 Hy tnh gi tr ca biu thc: P = a2004 + b2004.

Bi tp 2: Cho 1 1 1

+ + = 0.a b c

Tnh gi tr ca biu thc: 2 2 2

ab bc caS = + +

c a b.

Bi tp 3: Cho a3 + b3 + c3 = 3abc.

Tnh gi tr ca biu thc: a b c

S = 1 + 1 + 1 +b c a

Bi tp 4: Cho a + b + c = 0 v a2 + b2 + c2 = 14. Tnh gi tr ca biu thc: A = a4 + b4 + c4. Bi tp 5: Cho x + y + z = 0 v xy + yz + zx = 0. Tnh gi tr ca biu thc: B = (x - 1)

2007 + y

2008 + (z + 1)

2009.

Bi tp 6: Cho a2 + b2 + c2 = a3 + b3 + c3 = 1. Tnh gi tr ca biu thc: C = a2 + b9 + c1945. Bi tp 7: Tnh gi tr ca biu thc: a) A = x

4 - 17x

3 + 17x

2 - 1x + 20 ti x = 16.

b) B = x5 - 15x

4 + 16x

3 - 29x

2 + 13x ti x = 14.

c) C = x14

- 10x13

+ 10x12

- 10x11

+ ... + 10x2 - 10x + 10 ti x = 9.

d) D = x15

- 8x14

+ 8x13

- 8x12

+ ... - 8x2 + 8x - 5 ti x = 7.

Bi tp 8: Tnh gi tr ca biu thc: a) A = x

3(x

2 - y

2) + y

2(x

3 - y

3) vi x = 2; |y| = 1. b) B = M.N vi |x| = 2. Bit rng: M = -2x2 + 3x + 5; N = x2 - x + 3 Bi tp 9: Cho a3 - 3ab2 = 19 v b3 - 3a2b = 98. Hy tnh: E = a2 + b2.

Bi tp 10: Tnh gi tr ca biu thc: 81 1 36 1+ 14641 1- 14641

A = + + + +121 121 81 122 120

p s: A = 52

33

Bi tp 11: Tnh gi tr ca biu thc: N = 1296

243 + 363 + - 30723

Bi tp 12: Tnh gi tr ca biu thc: Q = 3 1x3x3xx21x , ti x = 5

p s: Q = 5 +1

Bi tp 13: Tnh gi tr ca biu thc sau: T = 50329872 p s: T = 0.

Bi tp 14: Cho biu thc sau: P = 35 + 2 17 5 -38

5 + 14 - 6 5 p s: P =

3

1

Bi tp 15: Tnh gi tr ca biu thc: G = 336 6216216425

p s: G = 0.



Bi tp 16: Tnh gi tr ca biu thc: E = 33

3

3

12.12

p s: E = 1.

Bi tp 17: Tnh gi tr ca biu thc: M = 526526 . p s: M = 2

Bi tp 18: Tnh gi tr ca biu thc: L = n2n

n2n

1x2x

x4x4

ti x = 2. p s: L = 2.

Bi tp 19: Tnh gi tr ca biu thc: Q = nn 2

n

x

2

x

11.x ti x = 2n. p s: Q = 3.

Bi tp 20: Tnh gi tr ca biu thc sau: A = 3 3 3 364 + 27 + 2 + 64 - 27 -1

p s: A = 3.

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CHUYN 10 TM GI TR NGUYN CA BIU THC

1. Kin thc c bn: Phng php:

tm gi tr x nguyn sao cho biu thc

A x

B x

(trong A(x), B(x) l nhng a thc) nhn gi

tr nguyn, ta lm nh sau:

Bin i

A x Q x

P x

B x B x

, Deg Q(x) < Deg B(x). Dn n tm s nguyn a sao cho Q(a)

chia ht cho B(a).

2. Bi tp p dng:

Bi tp 1: Tm s nguyn n sao cho biu thc

2n 5

n 2

l s nguyn.

Gii

Ta c:

2n 5 1

2

n 2 n 2

biu thc

2n 5

n 2

l s nguyn th

1

n 2

phi l s nguyn. V 2 l s nguyn.

Do : n + 2 phi l c ca 1. Suy ra:

n + 2 = 1 n = -1.

n + 2 = -1 n = -3.


6

n 6n 10

n 3

l s nguyn.

Gii

Ta c:

2

2 n 3 2n 6n 11 2n 3

n 3 n 3 n 3

Biu thc trn c gi tr nguyn khi

2

n 3

l s nguyn.

Do : n + 3 phi l c ca 2. Hay n + 3 = 2; n + 3 = -2; n + 3 = 1; n + 3 = -1

Vy n = -1; -5; -2; - 4.

Bi tp 3: Tm s nguyn x sao cho biu thc

5

3

x 1

x 1

nhn gi tr nguyn.

Gii

Ta c:

2 3 25 2

2 2

3 3 3 2

x x 1 x 1x 1 x 1 x 1

x x

x 1 x 1 x 1 x x 1

Biu thc

5

3

x 1

x 1

nhn gi tr nguyn khi

2x 1

x x 1

nhn gi tr nguyn vi x nguyn.

Suy ra:

2 2

x x 1 11

x x 1 x x 1

cng nhn gi tr nguyn.

(V mt s nguyn th nhn thm mt s nguyn x l mt s nguyn). Do :

Xt: 2x x 1 = 1 x(x - 1) = 0 hay x = 0 hoc x = 1.

Xt: 2

x x 1 = - 1 x2 - x + 2 = 0 (Phng trnh ny v nghim).



Vy gi tr x nguyn cn tm l x = 0; 1 tha mn yu cu bi ton. Bi tp 4: Tm k nguyn dng ln nht ta c s:

2

k 1

n

k 23

l mt s nguyn dng. Gii Ta c th vit:

2

2k 1 k 23 k 21 484k 2k 1 484n =k - 21+ , k Z

k 23 k 23 k 23 k 23

n l mt s nguyn khi v ch khi k + 23 l c ca 484, k + 23 > 23 Ta c: 484 = 22

2 = 4.121 = 44.11.

k 23 121 k 98

k 23 44 k 21

Vi k = 98, ta c: n = 81 Vi k = 21, ta c n = 11. Vy c hai gi tr k nguyn dng tha mn yu cu ca bi ton l k = 98 v k = 21. Gi tr ln nht ca k l 98.

Bi tp 5: Tm s nguyn z sao cho 1 1 1

2z 3z 4z

l mt s nguyn.

Gii

Ta c:

1 1 1 6 4 3 13

2z 3z 4z 12z 12z

biu thc trn l mt s nguyn th 13

12z

phi l mt s nguyn.

Hay 12z| 13. M 13 = 12.z v z Z. Nn khng tn ti v (12; 13) = 1.

Vy khng tn ti z sao cho 1 1 1

2z 3z 4z

l mt s nguyn.

3. Bi tp t luyn:


6n 13

2n 1

l mt s nguyn.

Bi tp 2: Tm s nguyn m n sao cho biu thc

2

n 6n 10

2n 9

l mt s nguyn m.

Bi tp 3: Tm s nguyn k nh nht sao cho biu thc

2

3k 2k 1(2k 1)

3k 1

t gi tr nguyn.

Bi tp 4: Tm s nguyn dng n sao cho biu thc 2n 1 2n 8 c gi tr nguyn.

Bi tp 5: Tm s nguyn m sao cho biu thc m 1 m 1 t gi tr nguyn.

Bi tp 6: Cho a = *1

x ,x R

x

, l mt s nguyn. Chng minh rng s 20052005

1b x

x

l mt

s nguyn.

Bi tp 7: Cho x v y l hai s thc khc 0 sao cho cc s: 1 1

a = x + ; b = y +y x

u l s

nguyn.

a) Chng minh rng s 2 22 2

1c = x y +

x y cng l mt s nguyn.

b) Tm mi s nguyn dng x, y sao cho s n nn n

1d x y

x y

l s nguyn.

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CHUYN 10 TM GI TR NGUYN CA BIU THC

1. Kin thc c bn: Phng php:

tm gi tr x nguyn sao cho biu thc

A x

B x

(trong A(x), B(x) l nhng a thc) nhn gi

tr nguyn, ta lm nh sau:

Bin i

A x Q x

P x

B x B x

, Deg Q(x) < Deg B(x). Dn n tm s nguyn a sao cho Q(a)

chia ht cho B(a).

2. Bi tp p dng:


2n 5

n 2

l s nguyn.

Gii

Ta c:

2n 5 1

2

n 2 n 2

biu thc

2n 5

n 2

l s nguyn th

1

n 2

phi l s nguyn. V 2 l s nguyn.

Do : n + 2 phi l c ca 1. Suy ra:

n + 2 = 1 n = -1.

n + 2 = -1 n = -3.


6

n 6n 10

n 3

l s nguyn.

Gii

Ta c:

2

2 n 3 2n 6n 11 2n 3

n 3 n 3 n 3

Biu thc trn c gi tr nguyn khi

2

n 3

l s nguyn.

Do : n + 3 phi l c ca 2. Hay n + 3 = 2; n + 3 = -2; n + 3 = 1; n + 3 = -1

Vy n = -1; -5; -2; - 4.

Bi tp 3: Tm s nguyn x sao cho biu thc

5

3

x 1

x 1

nhn gi tr nguyn.

Gii

Ta c:

2 3 25 2

2 2

3 3 3 2

x x 1 x 1x 1 x 1 x 1

x x

x 1 x 1 x 1 x x 1

Biu thc

5

3

x 1

x 1

nhn gi tr nguyn khi

2x 1

x x 1

nhn gi tr nguyn vi x nguyn.

Suy ra:

2 2

x x 1 11

x x 1 x x 1

cng nhn gi tr nguyn.

(V mt s nguyn th nhn thm mt s nguyn x l mt s nguyn). Do :

Xt: 2x x 1 = 1 x(x - 1) = 0 hay x = 0 hoc x = 1.

Xt: 2

x x 1 = - 1 x2 - x + 2 = 0 (Phng trnh ny v nghim).

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Vy gi tr x nguyn cn tm l x = 0; 1 tha mn yu cu bi ton. Bi tp 4: Tm k nguyn dng ln nht ta c s:

2

k 1

n

k 23

l mt s nguyn dng. Gii Ta c th vit:

2

2k 1 k 23 k 21 484k 2k 1 484n =k - 21+ , k Z

k 23 k 23 k 23 k 23

n l mt s nguyn khi v ch khi k + 23 l c ca 484, k + 23 > 23 Ta c: 484 = 22

2 = 4.121 = 44.11.

k 23 121 k 98

k 23 44 k 21

Vi k = 98, ta c: n = 81 Vi k = 21, ta c n = 11. Vy c hai gi tr k nguyn dng tha mn yu cu ca bi ton l k = 98 v k = 21. Gi tr ln nht ca k l 98.

Bi tp 5: Tm s nguyn z sao cho 1 1 1

2z 3z 4z

l mt s nguyn.

Gii

Ta c:

1 1 1 6 4 3 13

2z 3z 4z 12z 12z

biu thc trn l mt s nguyn th 13

12z

phi l mt s nguyn.

Hay 12z| 13. M 13 = 12.z v z Z. Nn khng tn ti v (12; 13) = 1.

Vy khng tn ti z sao cho 1 1 1

2z 3z 4z

l mt s nguyn.

3. Bi tp t luyn:


6n 13

2n 1

l mt s nguyn.

Bi tp 2: Tm s nguyn m n sao cho biu thc

2

n 6n 10

2n 9

l mt s nguyn m.

Bi tp 3: Tm s nguyn k nh nht sao cho biu thc

2

3k 2k 1(2k 1)

3k 1

t gi tr nguyn.

Bi tp 4: Tm s nguyn dng n sao cho biu thc 2n 1 2n 8 c gi tr nguyn.

Bi tp 5: Tm s nguyn m sao cho biu thc m 1 m 1 t gi tr nguyn.

Bi tp 6: Cho a = *1

x ,x R

x

, l mt s nguyn. Chng minh rng s 20052005

1b x

x

l mt

s nguyn.

Bi tp 7: Cho x v y l hai s thc khc 0 sao cho cc s: 1 1

a = x + ; b = y +y x

u l s

nguyn.

a) Chng minh rng s 2 22 2

1c = x y +

x y cng l mt s nguyn.

b) Tm mi s nguyn dng x, y sao cho s n nn n

1d x y

x y

l s nguyn.



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CHUYN 11 SO SNH MT S VI CC NGHIM CA PHNG TRNH BC HAI

1. Kin thc c bn: Gi s phng trnh bc hai: f(x) = ax2 + bx + c = a(x - x1)(x - x2) = 0 c hai nghim x1, x2, (a 0).

Theo nh l Vi-t, ta c:

1 2

1 2

bS = x + x = -

a

cP = x x =

a

Dng ton:

Dng 1: Tn ti s tha mn x1 < < x2.

Bin i: x1 - < 0 < x2 - .

t: x - = y x = y + .

Suy ra: f(y) = a(y + )2 + b(y + ) + c = 0 (*) tha mn yu cu bi ton th phng trnh (*) phi c hai nghim tri du.

P < 0.

Dng 2: Tn ti s tha mn < x1 < x2.

Ta c: < x1 < x2 0 < - x1 < - x2.

t: y = - x x = - y.

Chuyen de Daiso on Thi Lop10

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