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Trng THCS Nguyn Tri Phng GVGD: Trng Trng Hu

Nng Cao Ton 8

CHUYN 1: PHN TCH A THC THNH NHN T 1) Phn tch a thc thnh nhn t l bin i a thc thnh tch cc a thc.

2) Cc phng php thng dng : t nhn t chung, dng hng ng thc, nhm cc hng t v phi

hp c ba phng php.

3) Ngoi nhng phng php trn, vn cn mt s phng php, dng ton m rng Phn tch a thc

thnh nhn t nm ngoi chng trnh lp 8. Cc dng ton m rng u da trn nn tng ca cc

phng php c bn i ra.

Dng 1 : i Xng Vng Quanh Du hiu : Thay a = b ;b = c ; c = a : biu thc khng thay i.

Phng php : Thm bt phn t th 3 vo mt a thc bt k ca biu thc.

Bi mu : Phn tch a thc thnh nhn t :

them

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

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

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

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

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

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

(a b)(a c)(b c

a

)

a

Bi tp : Phn tch a thc thnh nhn t

2 2 2 3 3 3 3 3 3

1) xy(x y) yz(y z) zx(z x) 2) ab(a b) bc(b c) ac(a c)

3) a (b c) b (a c) c (a b) 4) a(b c ) b(c a ) c(a b )

5) ab(a b) bc(b c) ca(c a) 2abc 6) ab(b a) bc(b c) ca(c a) 2abc

Dng 2 : t n ph Du hiu : phn chung, ging nhau

Phng php : t t = phn chung , nguyn tc mn phi tr .

Bi mu : Phn tch a thc thnh nhn t

1) A = 6x4 11x2 + 3

= 6(x2)2 11x2 + 3

t t = x2

A= 6t2 11t + 3

= 6t2 2t 9t + 3

= 2t(3t 1) 3(3t 1)

= (3t 1)(2t 3)

= (3x2 1)(2x2 3)

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

t t = x2 x

Thay vo B = (t + 3)(t 2) + 4

= t2 + 3t 2t 6 + 4

= t2 + t 2

= t2 t + 2t 2

= t(t 1) + 2(t 1)

= (t 1)(t + 2)

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


Nng Cao Ton 8


4 2 4 2

2 2 2 2 2 2

2 2 2 2

1) x 3x 4 2) 4x 37x 9

3) (x x) 4(x x) 12 4) (x 3x) 7x 21x 10

5) (x x 1)(x x 2) 12 6) (2x x 1)(2 x x 4) 10

Dng 3 : Bin ph hi quy

4 3 2A ax bx cx dx e

Du hiu : 2d e

m; mb a

Phng php : t m

t xx

Bi mu : Phn tch a thc thnh nhn t 4 3 2

2 4 3 2

2

4 3 22

2

2 2

2

2 2

2

2 2

2

B

A x 6x 11x 6x 1

x (x 6x 11x 6x 1)

x

x 6x 11x 6x 1x .

x

6 1x . x 6x 11

x x

1 6x . x 6x 11

x x

1 1x . x 6 x 11

x x

t 1

t xx

2 2 2

2 2

2 2

2

1 1 1t x 2.x. x 2

x x x

1t 2 x

x

Thay vo B

2 2

2 2

2

B (t 2) 6.t 11 t 2 6t 11

t 6t 9 (t 3)

1x 3

x

Vy

2

2 1A x x 3x

Bi tp : Phn tch a thc thnh nhn t 4 3 2 4 3 2

4 3 2 4 3 2

1)x x 4x x 1 2)x 5x 12x 5x 1

3)2x 5x 27x 25x 50 4)4x 6x 33x 24x 56


Nng Cao Ton 8

Dng 4 : Phn tch a thc da vo HT A2 B2 ( qui v bi ton tm GTLN, GTNN )

Du hiu : 2Ax Bx C

Phng php : Tch v HT 2 2(A B) C


2

2 2

2 2

2 2

A 2AB2.A.B

+B B

2 2

2 2

2 2 2

5 5 5a) x 5x 6 x 2.x. 6

2 2 2

5 25 5 49x 6 x

2 4 2 4

5 7 5 7 5 7x x x

2 2 2 2 2 2

x 1 x 6

5 3 5 5b)2x 5x 3 2 x x 2 x 2.x.

2 2 4 4

2 2

2

5 3

4 2

5 1 5 1 5 12 x 2 x x

4 16 4 4 4 4

32 x 1 x

2


2 2 2

2 2 2

2

a)x 4x 3 b)x 5x 4 c)x x 6

d)2x 7x 3 e)3x 10x 8 f )3x 7x 76

1 19g) x x 1

2 6

Dng 5 : Thm bt cng mt s hng 1) Thm bt xut hin A2 B2

Bi mu :

2 24 2 2 2 2 2 2

2.A.B 2.A.B

2 22 2 2

x 64 x 8 x 8 16x 16x

x 8 4x x 4x 8 x 4x 8

2) Thm bt cng mt s hng xut hin nhn t chung :

Bi mu :

5 5 2 2 2 3 2

Nhan Tu Chung

2 2 2

2 3 2

x x 1 x x x x 1 x x 1 x x 1

x (x 1)(x x 1) x x 1

x x 1 x x 1

Lu :

Cac a thc dang: x3m+2

+ x3n+1

+ 1 luon cha tha so x2 + x + 1


Nng Cao Ton 8

Bi tp : Phn tch a thc thnh nhn t 4 4 4

4 2 12 6 16 8

5 7 2 5 4

7 10 5

a) x 4 b)x 324 c)x 64

d)x x 1 e)x x 1 f ) x x 1

g)x x 1 h)x x 1 i)x x 1

j)x x 1 g)x x 1

Dng 6 : t n ph dng (x a)(x b)(x c)(x d) e

Du hiu : a c b d

Phng php : (x a)(x c)(x b)(x d) e nhn tng cm li, t n ph.


2 2

A (x 1)(x 2)(x 3)(x 4) 80

=(x 1)(x 4)(x 2)(x 3) 80

=(x 5x 4)(x 5x 6) 80

Nhn xt :a 1;b 2;c 3;d 4

a d b c 5

t 2t x 5x 5

2 2

2 2

2 2

A (t 1)(t 1) 80 t 1 80 t 81

= t 9 t 9 x 5x 5 9 x 5x 5 9

x 5x 4 x 5x 14


a) x 2 x 4 x 6 x 8 16 b) x 2 x 3 x 4 x 5 24

c)x x 4 x 6 x 10 128 d) x 1 x 3 x 5 x 7 20

e) x 1 x 2 x 3 x 6 28 f ) x 2 x 3 x 7 x 8 144

Dng 7 : t n ph dng 4 4

x a x b

Du hiu :

Phng php : t a b

t x2

S dng hng ng thc : 4 4 3 2 2 3 4a b a 4a b 6a b 4ab b


4 4

B x 3 x 5

t 3 5

t x x 42

4 4 4 3 2 4 3 2

4 2

B t 1 t 1 = t 4t 6t 4t 1 t 4t 6t 4t 1

= 2t 12t 2

Quay v bi ton t 2t x (dng 2)


4 4 4 4

4 4 4 4

a) x 2 x 8 b) x 6 x 4

c) x 2 x 4 d) x 8 x 6


Nng Cao Ton 8

CHUYN 2: NH L BEZOUT V S HORNER

PHP CHIA A THC :

nh Ngha :

Php chia a thc f(x) cho a thc g(x) ( khc 0) ta c thng l q(x) v d l r(x).

Ta vit : f(x) = q(x).g(x) + r(x) vi bc r(x) < bc g(x).

Nu a thc r(x) bng 0, ta c f(x) = q(x).g(x). Khi , ta ni : a thc f(x) chia ht

cho a thc g(x).

1. nh L BEZOUT (nh l Bedu):

S d trong php chia a thc f(x) cho a thc x a l f(a).

p dng :

a) Tm s d ca php chia a thc f(x) cho a thc g(x). Bit

4 3 23x 4x 12x 24x 1f(x)

g(x) x 2

Bi lm

Tm x khi g(x) 0 x 2

p dng nh l Bedu :

S d ca php chia a thc f(x) cho a thc g(x) l f(2)

f(2) 79

b) Tm d ca php chia a thc 5 4 3 236x x 27x 1

Px 2

5x

3

( d l

5587

16)

2. S HOCNER: Xt a thc f(x) chia cho x a .

Theo Bedu th ta bit s d a thc f(x) chia cho x a l f(a).

Hay c th ni cch khc l : f(x) = (x a).q(x) + f(a).

Vn y l ta lm th no xc nh c thng ca php chia a thc. Ni cch

khc l xc nh q(x).

By gi, ta s lm quen vi mt thut ton tm thng q(x). Ta gi l Thut Ton

Horner (s Hocne).

Thut Ton Horner :

Xt a thc: 5 4 3 2

0 1 2 3 4 5f (x) a x a x a x a x a x a v x = k l nghim ca a thc th

Ta c s Hocner: (Quy tc : Nhn ngang cng cho )

Lc : 4 3 2

0 0 1 2 3f (x) (x k)(a x b x b x b x b ) (x k).q(x)

Ngoi ra, s Hocne cn mt s cng dng khc :

+ H bc mt a thc xung 1 bc.

+ Tm s d ca php chia a thc: f(x) chia cho (x k) c s d l b4

a0 a1 a2 a3 a4 a5

x = k a0 a0.k + a1

= b0

b0.k + a2

= b1

b1.k + a3

= b2

b2.k + a4

= b3

b3.k + a5

= b4


Nng Cao Ton 8

+ Tm h s ca thng hai a thc: f(x) chia cho (x k) c s h s l a0 ;b0 ;b1 ;b2 ; b3

+ Phn tch a thc thnh nhn t bng s Hocne :

Phng Php :

Bc 1 : Chn mt gi tr x = a l mt nghim ca a thc f(x).

Bc 2 : S dng s Hocne a a thc f(x) = (x a).q(x).

Bc 3 : Thc hin ging nh bc 1 i vi a thc q(x).

Chn 1 gi tr x = b l nghim ca a thc q(x). Sau tip tc dng s

Hocne. Ta s phn tch c thnh f(x) = (x a).q(x) = (x a)(x b).h(x)

Bc 4 : Tip tc phn tch thnh nhn t h(x) nu nh cn phn tch c.

p dng :

1) Phn tch a thc 4 3f(x) 3x 4x 1 thnh nhn t.

Bi lm

Nhn xt : x = 1 l 1 nghim ca a thc :

S Hocne :

3 4 0 0 1

1 3 3.1 ( 4) 1 ( 1).1 0 1 ( 1).1 0 1 ( 1).1 1 0

T y, h s cc a thc thng q(x) ln lt l : 3; 1; 1; 1 v c d l 0

Lc 3 2f(x) (x 1)(3x x x 1) (x 1).q(x)

Tip tc, ta c x = 1 l nghim ca a thc q(x)

S Hocne :

3 -1 -1 -1

1 3 3.1 ( 1) 2 2.1 ( 1) 1 1.1 ( 1) 0

Lc 2q(x) (x 1).(3x 2x 1)

Suy ra : 2 2 2f(x) (x 1).(x 1).(3x 2x 1) (x 1) (3x 2x 1)

2) Phn tch a thc thnh nhn t : 3 2x) x 6h x 4x(

3. nh l v nghim nguyn ca a thc :

a. Tiu chun phn tch ca mt a thc.

Xt trn a thc bc n :

Cho a thc n n 1 n 2 2

n n 1 n 2 2 1 0 nf (x) a x a x a x ... a x a x a (a 0)

c n nghim 1 2 3 n

b ;b ;b ;...;b

Theo tiu chun phn tch a thc:

n 1 2 3 n

f(x) a (x b )(x b )(x b )...(x b )

V d:

Cho a thc: 2f(x) 3x 5x 2

Nhn xt: 2

x 1;x3

l nghim ca a thc.

Nn a thc 2

f(x) 3(x 1) x3


Nng Cao Ton 8

b. nh l v nghim nguyn ca a thc :

Cho a thc n n 1 n 2 2n n 1 n 2 2 1 0 nf (x) a x a x a x ... a x a x a (a 0)

Nu p

q l mt nghim ca a thc f(x), th p l c ca

0a v q l c ca

na

H qu :

Nu f(x) c nghim nguyn th nghim nguyn phi l c ca 0

a

Khi n

a 1 , a thc n n 1 n 2 2

n 1 n 2 2 1 0 nf (x) x a x a x ... a x a x a (a 0)

th nghim ca a thc f(x) s l c ca 0

a .

Nh vy, ta c th s dng nh l v nghim ca a thc kt hp vi s Hocne phn tch a thc

thnh nhn t hay c th tm c tt c cc nghim cn li ca mt a thc.

p dng :

Vn dng nh l v nghim ca a thc, nh l Bedu, s Hocne phn tch a thc thnh nhn t :

4 3 2f(x) x 2x 4x 5x 6

Bi lm

- Nghim ca a thc f(x) (nu c) phi l c ca (-6)

- Ln lt th cc c ca (-6). Ta thy f(2) 0;f( 3) 0 . Nn x 2;x 3 l 2 nghim ca f(x)

- Lc : f(x) (x 2)(x 3).q(x)

- S dng s Hocne tm a thc q(x)

1 2 4 5 -6

2 1 4 4 3 0

3 1 1 1 0

- Xc nh c 2q(x) x x 1

- Khi : 2f(x) (x 2)(x 3)(x x 1)

- M

2

2 1 3q(x) x x 1 x 02 4

x . Nn a thc q(x) v nghim.

- Vy 2f(x) (x 2)(x 3)(x x 1)

BI TP

1. Vn dng nh l Bedu. Tm d ca php chia a thc

7 5 3

9 7 4 2

4 3

4x 12x 3x x 1

3x 5x x x 201x 2014

2

A 275371x 5

4417B

x 1

1x 12x x

x

2C3

5

2

2


Nng Cao Ton 8

2. Tm a a thc f(x) chia ht cho a thc g(x), vi:

a) 4 3 2f(x) x 9x 21x ax 2 , g(x) x 2

b) 4 3 2f(x) x x 6x x a , g(x) x 5 (a 645)

c) 3 2f(x) 3x 10x 5 a , g(x) 3x 1 (a 4)

d) 3f(x) x 3x a , g(x) 3x2 46

a27

3. Vn dng nh l v nghim a thc, nh l Bedu, s Hocne, phn tch a thc thnh nhn t : 3 2

3

4 3 2

6 4 2

3 2

3 2

3 2

a(x) x 6x 13x 42

b(x) x 7x 6

c(x) x 8x 10x 104x 105

d(x) x 14x 49x 36

e(x) x x 8x 12

f(x) 3x 5x 14x 4

g(x) 2x x 3x 1

4. Vn dng tiu chun phn tch ca a thc, phn tch a thc sau thnh nhn t :

2 2 2 3 3 3 3 3 3

1) xy(x y) yz(y z) zx(z x) 2) ab(a b) bc(b c) ac(a c)

3) a (b c) b (a c) c (a b) 4) a(b c ) b(c a ) c(a b )

5) ab(a b) bc(b c) ca(c a) 2abc 6) ab(b a) bc(b c) ca(c a) 2abc

5. Xc nh hng s a,b a thc 3h(x) 3x ax b chia (x + 1) d 6 ; chia (x 3) d 70

6. Xc nh hng s a,b a thc 3m(x) 2x ax b chia (x + 1) d -6 ; chia (x 2) d 21

7. Tm a v b : 3 2

4 3 2 2

a)f(x) x ax b chia ht cho x 5x 6

b)g(x) x x 5x ax b chia cho x x 2 d 2x + 1

K TRC : 4 2 2 2

12 6 6 3 6 3

16 8

5

7 2

5 4

7

2 2 4 2 8 4

2 3 2

2 5 4 2

2 3

2

d)x x 1 (x x 1)(x x 1)

e)x x 1 (x x 1)(x x 1)

f ) x x 1

g

(x x 1)(x x 1)(x x 1)(x x 1)

(x x 1)(x x 1)

(x x 1)(x x x

)x x 1

h)x x 1

i)x x 1

j

x 1)

(x x 1)(x x 1)

(x) xx x 1

5 4 3

10 5 2 8 7 5 4 3

1)(x x x x 1)

(x x 1)(xk)x x 1 x x x x x 1)


Nng Cao Ton 8

CHUYN 4: PHNG TRNH PHNG TRNH CHA DU GI TR TUYT I

PHNG TRNH

Bi 1: Gii cc phng trnh sau : 3 2 3 2

3 2 3 2

4 2 2 2

a)x 2x x 2 0 b)x 2x x 2 0

c)x x 21x 45 0 d)x 3x 4x 2 0

e)x x 6x 8 0 f)(x 1) 4(2x 1)

Bi 2: Gii cc phng trnh : 4 2

4 2

2 2 2

2 2 2

2 2

2

2 2 2

2 2 4 2

2

a)x 3x 4 0

b)4x 37x 9 0

c)(x 5x) 2(x 5x) 24 0

d)(x 5x) 10(x 5x) 24 0

e)(x x 1)(x x 2) 12

f )x(x 1)(x x 1) 42

g)(x 6x 9) 15(x 6x 10) 1

h)(x x 1) 3(x x 1)

i)2x(8x 1) (4x 1) 9

Bi 3: Gii cc phng trnh :

a)x(x 1)(x 1)(x 2) 24

b)(x 4)(x 5)(x 6)(x 7) 1680

c)(x 2)(x 3)(x 5)(x 6) 180


4 4

4 4

4 4

4 4

a)(x 3) (x 5) 16

b)(x 2) (x 3) 1

c)(x 1) (x 3) 82

d)(x 2,5) (x 1,5) 1


4 3 2

4 3 2

4 3 2

4 3 2

4 3 2

a)x 3x 4x 3x 1 0

b)3x 13x 16x 13x 3 0

c)6x 5x 38x 5x 6 0

d)6 x 7x 36x 7x 6 0

e)2x 9x 14x 9x 2 0


Nng Cao Ton 8

PHNG TRINH CHA DU GIA TR TUYT DI :

gii cc phng trnh cha n trong du gi tr tuyt i, ta cn phi kh du gi tr tuyt

i. Nh li :

A nu A 0

AA nu A 0

Do , kh du gi tr tuyt i, ta cn xt gi tr ca bin lm sao cho biu thc trong du

gi tr tuyt i khng m hay m.

Nu biu thc trong du gi tr tuyt i l nh thc bc nht, ta cn nh n nh l sau :

nh l v du ca nh thc bc nht : ax b (a 0)

Gi x0 l nghim ca nh thc bc nht ax + b 0b

xa

Bng xt du nh thc bc nht :

x b

a

ax + b tri du vi a 0 cng du vi a

p dng:

Bi 1 : Gii phng trnh

2x 1 3x 2 (1)

1x

2 l nghim ca nh thc 2x 1

Lp bng xt du ca nh thc 2x 1

Trng Hp 1 : nu 1

x2

Phng trnh (1) 2x 1 3x 2

2x 1 3x 2

5 x (loi)

3

Trng Hp 2 : nu 1

x2

Phng trnh (1) 2x 1 3x 2

2x 1 3x 2

x 1 (nhn)

Vy phng trnh c tp hp nghim S 1

x 1

2

2x 1 + 0


Nng Cao Ton 8

Bi 2 : Gii phng trnh :

x 5 x 3 3x 1 (2)

x = 5 ; x = 3 l nghim ca nh thc x 5 v x + 3

Lp bng xt du ca nh thc x 5 v x + 3

Trng Hp 1 : Nu x 3

Phng trnh (2) x 5 x 3 3x 1

(5 x) (x 3) 3x 1

3x = (loi)

5

Trng Hp 2 : Nu 3 x 5


(5 x) (x 3) 3x 1

x = 3 (nhn)

Trng Hp 3 : Nu x 5


x 5 x 3 3x 1

x = 1 (loi)

Vy phng trnh c tp hp nghim S 3

Bi tp:

Bi 1: Gii cc phng trnh:

a) 2x 1 x 3 b) 3x 5 2x 7

1 3c) 3 x x 6 d) 5x 2 2x

2 2

Bi 2 : Gii cc phng trnh :

a)2 x 3 (5x 1) 0 b) x 1 x 5

c) x 1 3x 5 d)2 x x 1 2

e) x 4 x 1 9 f) x 3 x 5 2

Bi 3 : Gii cc phng trnh :

3 2

a) x 2 x 2 3 x 3 4 b) x 2 x 1 3 x 2 0

c) x 2 x 1 3 x 2 0 d) x 1 x 2 x 3 4x

e) x 3 x 2 0 f)x x 3 x x 1 1

x 3 5

x 5 | 0 +

x + 3 0 + | +

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