a) \(=\left(2x-1\right)^2\)

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

a. \(4x^2-4x+1=\left(2x\right)^2-2x.2.1+1^2=\left(2x-1\right)^2\)

b. \(xy^2-x^3+2x^2-x=x\left(y^2-x^2+2x-1\right)=x\left[y^2-\left(x^2-2x+1\right)\right]=x\left[y^2-\left(x-1\right)^2\right]=x\left(y-x+1\right)\left(y+x-1\right)\)

Cưu là mình vs (x^2+x)^2-2(x^2+x)-15

a) \(36a^4-y^2=\left(6a^2-y\right)\left(6a^2+y\right)\)

b) \(6x^2+x-2=2x\left(3x+2\right)-1\left(3x+2\right)=\left(3x+2\right)\left(2x-1\right)\)

\(a,x-xy+y-y^2\\=(x-xy)+(y-y^2)\\=x(1-y)+y(1-y)\\=(1-y)(x+y)\\---\\b,x^2-4x-y+4(?)\\---\\c,x^2-2x-3\\=x^2+x-3x-3\\=x(x+1)-3(x+1)\\=(x+1)(x-3)\)

Bạn xem lại đề câu b nhé!

`x-xy+y-y^2`

`=x(1-y)+y(1-y)`

`=(1-y)(x+y)`

__

`x^2-4x-y+4`

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

`= (x-2)^2-y`

Thiếu đề?

__

`x^2-2x-3`

`=x^2+x-3x-3`

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

`=x(x+1)-3(x+1)`

`=(x+1)(x-3)`

a

\(8x^3-\dfrac{1}{125}y^3\\ =\left(2x\right)^3-\left(\dfrac{1}{5}y\right)^3\\ =\left(2x-\dfrac{1}{5}y\right)\left[\left(2x\right)^2+2x.\dfrac{1}{5}y+\left(\dfrac{1}{5}y\right)^2\right]\\ =\left(2x-\dfrac{1}{5}y\right)\left(4x^2+\dfrac{2}{5}xy+\dfrac{1}{25}y^2\right)\)

b

\(-x^3+6x^2y-12xy^2+8y^3\\ =-\left(x^3-6x^2y+12xy^2-8y^3\right)\\ =-\left(x^3-3.2y.x^2+3.\left(2y\right)^2.x-\left(2y\right)^3\right)\\ =-\left(x-2y\right)^3\\ =-\left(x-2y\right)\left(x-2y\right)\left(x-2y\right)\)

a: 8x^3-1/125y^3

=(2x)^3-(1/5y)^3

=(2x-1/5y)(4x^2+2/5xy+1/25y^2)

b: =(2y-x)^3

a) \(8x^3-\dfrac{1}{125}y^3\)

\(=\left(2x\right)^3-\left(\dfrac{1}{5}y\right)^3\)

\(=\left(2x-\dfrac{1}{5}y\right)\left[\left(2x\right)^2+2x\cdot\dfrac{1}{5}y+\left(\dfrac{1}{5}y\right)^2\right]\)

\(=\left(2x-\dfrac{1}{5}y\right)\left(4x^2+\dfrac{2}{5}xy+\dfrac{1}{24}y^2\right)\)

b) \(-x^3+6x^2y-12xy^2+8y^3\)

\(=-\left(x^3-6x^2y+12xy^2-8y^2\right)\)

\(=-\left(x^3-3\cdot x^2\cdot2y+3\cdot x\cdot\left(2y\right)^2-\left(2y\right)^3\right)\)

\(=-\left(x-2y\right)^3\)

a) \(=x^4-14x^2+40-72=x^4-14x^2-32=\left(x-4\right)\left(x+4\right)\left(x^2+2\right)\)

b) \(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1=\left(x^2+5x\right)^2+2\left(x^2+5x\right)+1=\left(x^2+5x+1\right)^2\)

c) \(=x^4+3x^3-3x^2+3x^3+9x^2-9x+x^2+3x-3-5=x^4+6x^3+7x^2-6x-8=\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x+4\right)\)

a: Ta có: \(\left(x^2-4\right)\left(x^2-10\right)-72\)

\(=x^4-14x^2-32\)

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

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

b: Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)

\(=\left(x^2+5x+6\right)\left(x^2+5x+4\right)+1\)

\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24+1\)

\(=\left(x^2+5x+1\right)^2\)

\(a,=\left(x-1\right)^4-2\left(x-1\right)^2+1\\ =\left[\left(x-1\right)^2-1\right]^2\\ =\left(x^2-2x-2\right)^2\\ b,=\left[\left(x+1\right)\left(x+5\right)\right]\left[\left(x+2\right)\left(x+4\right)\right]-4\\ =\left(x^2+6x+5\right)\left(x^2+6x+8\right)-4\\ =\left(x^2+6x\right)^2+13\left(x^2+6x\right)+36\\ =\left(x^2+6x+4\right)\left(x^2+6x+9\right)\\ =\left(x+3\right)^2\left(x^2+6x+4\right)\)

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

Lời giải:

a.

$=6x^3y^2(2-x)-9x^2y^2(2-x)$

$=(2-x)(6x^3y^2-9x^2y^2)$

$=(2-x).3x^2y^2(2x-3)=3x^2y^2(2-x)(2x-3)$

b.

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

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

c.

$81x^2-(9y^2-6yz+z^2)$

$=(9x)^2-(3y-z)^2=(9x-3y+z)(9x+3y-z)$

\(=\left(6x^2-3x\right)+\left(4x-2\right)\)

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

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

\(=6x^2-3x+4x-2=6x\left(x-2\right)+2\left(x-2\right)=2\left(3x+2\right)\left(x-2\right)\)