\(18a^3b^2-9a^2b^3\)

b) \(x^2...">

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18 tháng 11 2018

a) 18a^3b^2-9a^2b^3

=9a^2b^2(2a-b).

b) đề bài sai nha, phải là x^2-6xy+9y^2-36 nha

c) 2x^2-2xy-x+y

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

= (2x-1)(x-y).

d) x^2+6x-4y^2+9

= x^2+6x+9 -4y^2

= (x+3)^2- (2y)^2

= (x+3-2y)(x+3+2y).

Chắc chắn đúng 100% nha !!!

18 tháng 11 2018

a) 18a3b2−9a2b3=9a2b2(2a-b)

c)2x2−2xy−x+y=x(2x-1)-y(2x-1)=(2x-1)(x-y)

2 tháng 9 2018

\(x^3+8y^3+2xy^2+x^2y\)

\(=x^3+2x^2y-x^2y-2xy^2+4xy^2+8y^3\)

\(=x^2\left(x+2y\right)-xy\left(x+2y\right)+4y^2\left(x+2y\right)\)

\(=\left(x+2y\right)\left(x^2-xy+4y^2\right)\)

24 tháng 9 2017

Dài dữ trời :V Về sau gửi từng bài một thôi, nhìn hoa mắt quá @@

B1: Phân tích thành nhân tử:

a) \(6x^2+9x=3x\left(2x+3\right)\)

b) \(4x^2+8x=4x\left(x+2\right)\)

c) \(5x^2+10x=5x\left(x+2\right)\)

d) \(2x^2-8x=2x\left(x-4\right)\)

e) \(5x-15y=5\left(x-3y\right)\)

f) \(x\left(x^2-1\right)+3\left(x^2-1\right)=\left(x^2-1\right)\left(x+3\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(x+3\right)\)

g) \(x^2-2x+1-4y^2=\left(x-1\right)^2-4y^2\)

\(=\left(x-1-2y\right)\left(x-1+2y\right)\)

h) \(x^2-100=\left(x-10\right)\left(x+10\right)\)

i) \(9x^2-18x+9=\left(3x-3\right)^2\)

k) \(x^3-8=\left(x-2\right)\left(x^2+2x+4\right)\)

l) \(x^2+6xy^2+9y^4=\left(x+3y\right)^2\)

m) \(4xy-4x^2-y^2=-\left(4x^2-4xy+y^2\right)\)

\(=-\left(2x-y\right)^2\)

n) \(\left(x-15\right)^2-16=\left(x-15-16\right)\left(x-15+16\right)\)

\(=\left(x-31\right)\left(x+1\right)\)

o) \(25-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3+x\right)\)

\(=\left(2+x\right)\left(8+x\right)\)

p) \(\left(7x-4\right)^2-\left(2x+1\right)^2\)

\(=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)\)

\(=\left(5x-5\right)\left(9x-3\right)\)

24 tháng 9 2017

Bài 1 :

a ) \(6x^2+9x=3x\left(x+3\right)\)

b ) \(4x^2+8x=4x\left(x+2\right)\)

c ) \(5x^2+10x=5x\left(x+2\right)\)

d ) \(2x^2-8x=2x\left(x-4\right)\)

e ) \(5x-15y=5\left(x-3y\right)\)

f ) \(x\left(x^2-1\right)+3\left(x^2-1\right)=\left(x^2-1\right)\left(x+3\right)\)

g ) \(x^2-2x+1-4y^2=\left(x-1\right)^2-\left(2y\right)^2=\left(x-1-2y\right)\left(x-1+2y\right)\)

h ) \(x^2-100=x^2-10^2=\left(x-10\right)\left(x+10\right)\)

i ) \(9x^2-18x+9=\left(3x-3\right)^2\)

k ) \(x^3-8=\left(x-2\right)\left(x^2+2x+2^2\right)\)

l ) \(x^2+6xy^2+9y^4=\left(x+3y^2\right)^2\)

m ) \(4xy-4x^2-y^2=-\left(2x-y\right)^2\)

n ) \(\left(x-15\right)^2=x^2-30x+15^2\)

o ) \(25-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3-x\right)=\left(2+x\right)\left(8-x\right)\)

p ) \(\left(7x-4\right)^2-\left(2x+1\right)^2=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)=\left(5x-5\right)\left(9x-3\right)\)

Bài 2 :

a ) \(3x^3-6x^2+3x^2y-6xy=3x\left(x^2-2x+xy-2y\right)\)

b ) \(x^2-2x+xy-2y=x\left(x-2\right)+y\left(x-2\right)=\left(x-2\right)\left(x+y\right)\)

c ) \(2x+x^2-2y-2xy=......................\)

d ) \(x^2-2xy+y^2-9=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)

e ) \(x^2+y^2-2xy-4=\left(x-y\right)^2-2^2=\left(x-y-2\right)\left(x-y+2\right)\)

f )\(2xy-x^2-y^2+9=-\left(x-y\right)^2+9=3^2-\left(x-y\right)^2=\left(3-x+y\right)\left(3+x-y\right)\)

2 tháng 7 2017

a) \(25.\left(x-1\right)^2-16\left(x+y\right)^2\)

= \(\left(5x-5\right)^2-\left(4x+y\right)^2\)

= \(\left(5x-5-4x-y\right)\left(5x-5+4x+y\right)\)

= \(\left(x-y-5\right)\left(9x+y-5\right)\)

b) \(x^3+3x^2+3x+1-27z^3\)

= \(\left(x+1\right)^3-27z^3\)

= \(\left(x+1-3z\right)\left(x^2+x.3z+9z^2\right)\)

c) \(x^2-2xy+y^2-xz+yz\)

= \(\left(x-y\right)^2-z\left(x-y\right)\)

= \(\left(x-y\right)\left(x-y-z\right)\)

d) \(a^3x-ab+b-x\)

= \(x\left(a^3-1\right)-b\left(a-1\right)\)

= \(x\left(a-1\right)\left(a^2+a+1\right)-b\left(a-1\right)\)

= \(\left(a-1\right)\left(a^2x+ax+x-b\right)\)

f) \(x^2+2x-4y^2-4y\)

= \(x^2+2x+1-\left(4y^2+4y+1\right)\)

= \(\left(x+1\right)^2-\left(2y+1\right)^2\)

= \(\left(x+1-2y-1\right)\left(x+1+2y+1\right)\)

= \(\left(x-2y\right)\left(x+2y+2\right)\)

g) \(xy-4+2x-2y\)

= \(y\left(x-2\right)-2\left(x-2\right)\)

= \(\left(x-2\right)\left(y-2\right)\)

a: \(=\left(5x-5\right)^2-\left(4x-4y\right)^2\)

\(=\left(5x-5-4x+4y\right)\cdot\left(5x-5+4x-4y\right)\)

\(=\left(x+4y-5\right)\left(9x-4y-5\right)\)

b: \(=\left(x+1\right)^3-\left(3z\right)^3\)

\(=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)

c: \(=\left(x-y\right)^2-z\left(x-y\right)\)

\(=\left(x-y\right)\left(x-y-z\right)\)

d: \(=x\left(a^3-1\right)-b\left(a-1\right)\)

\(=x\left(a-1\right)\cdot\left(a^2+a+1\right)-b\left(a-1\right)\)

\(=\left(a-1\right)\left(a^2x+ax+1-b\right)\)

5 tháng 11 2017

1.

a. x2 - 2x + 1 = 0

x2 - 2x*1 + 12 = 0

(x-1)2 = 0

............( tới đây tui bí rùi tự suy nghĩ rùi lm tiếp ik)

1, Tìm x biết:

a, x2 - 2x +1 = 0

(x-1)2 = 0

x-1 = 0

x = 1. Vậy ...

b, ( 5x + 1)2 - (5x - 3) ( 5x + 3) = 30

25x2 +10x + 1 - (25x2 -9) = 30

25x2 +10x + 1 - 25x2 +9 = 30

10x + 10 =30

10(x+1) = 30

x+1 =3

x = 2. vậy ...

c, ( x - 1) ( x2 + x + 1) - x ( x +2 ) ( x - 2) = 5

(x3 - 1) - x(x2 -4) = 5

x3 - 1 - x3 + 4x = 5

4x - 1 = 5

4x = 6

x = \(\dfrac{3}{2}\) .vậy ...

d, ( x - 2)3 - ( x - 3) ( x2 + 3x + 9 ) + 6 ( x + 1)2 = 15

x3 - 6x2 + 12x - 8 - (x3 - 27) + 6 (x2 + 2x +1) =15

x3 - 6x2 + 12x - 8 - x3 + 27 + 6x2 + 12x +6 =15

24x + 25 = 15

24x = -10

x = \(\dfrac{-5}{12}\) vậy ...

6 tháng 3 2020

\(x^2+4x-y^2+4\\ =\left(x^2+4x+4\right)-y^2\\ =\left(x+2\right)^2-y^2\\ =\left(x+2-y\right)\cdot\left(x+2+y\right)\)

\(2xy-x^2-y^2+16\\ =\left(x^2-2xy+y^2\right)-16\\ =\left(x-y\right)^2-16\\ =\left(x-y+4\right)\cdot\left(x-y-4\right)\)

\(x^2-2x-4y^2-4y\\ =\left(x^2-4y^2\right)-\left(2x+4y\right)\\ =\left(x-2y\right)\cdot\left(x+2y\right)-2\left(x+2y\right)\\ =\left(x+2y\right)\cdot\left(x-2y+2\right)\)

\(x^2+6x+9-y^2\\ =\left(x-3\right)^2-y^2\\ =\left(x-3-y\right)\cdot\left(x-3+y\right)\)

\(3x^2+6xy+3y^2-3z^2\\ =3\cdot\left(x^2+2xy+y^2-z^2\right)\\ =3\cdot\left[\left(x^2+2xy+y^2\right)-y^2\right]\\ =3\cdot\left[\left(x-y\right)^2-z^2\right]\\ =3\cdot\left(x-y-z\right)\cdot\left(x-y+z\right)\)

\(9x-x^3\\ =x\cdot\left(9-x^2\right)\\ =x\cdot\left(3-x\right)\cdot\left(3+x\right)\)

\(\left(2xy+1\right)^2-\left(2x+y\right)^2\\ =\left(2xy+1-2x-y\right)\cdot\left(2xy+1+2x-y\right)\)

5 tháng 10 2020

a) x2-y2-2x+2y

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

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

b) 2x + 2y - x2 -xy

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

=(x+y)(2-x)

c) 3a2 - 6ab + 3b2 - 12c2

= 3(a2+b2) -3(2ab+4c2)

= 3(a2+b2-2ab-4c2)

d) x2 - 25 + y2 + 2xy

= x2 + 2xy + y2 -25

= (x+y)2 - 52

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

e) x2y - x3 - 9y + 9x

= (9x - x3)+(x2y -9y)

= x(9-x2)+y(x2 - 9)

= x(9-x2)-y(9-x2)

= (9-x2)(x-y)

f) x2-2x-4y2-4y

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

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

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

câu g trùng với câu e

h) x2(x-1)+16(1-x)

= x2(x-1)-16(x-1)

= (x2-16)(x-1)

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

1) \(x^2+6x+8\)

\(=x^2+2x+4x+8\)

\(=x\left(x+2\right)+4\left(x+2\right)\)

\(=\left(x+4\right)\left(x+2\right)\)

2) \(x^2-5x-14\)

\(=x^2-7x+2x-14\)

\(=x\left(x-7\right)+2\left(x-7\right)\)

\(=\left(x-7\right)\left(x+2\right)\)

3) \(2x^2+5x+3\)

\(=2x^2+2x+3x+3\)

\(=2x\left(x+1\right)+3\left(x+1\right)\)

\(=\left(x+1\right)\left(2x+3\right)\)

4) \(x^2-x-12\)

\(=x^2-4x+3x-12\)

\(=x\left(x-4\right)+3\left(x-4\right)\)

\(=\left(x-4\right)\left(x+3\right)\)

22 tháng 8 2018

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