\(a,=\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\\ b,=\left(x-\sqrt{3}y\right)\left(x+\sqrt{3}y\right)\\ c,=\left[3x-2y-2\left(x+y\right)\right]\left[3x-2y+2\left(x+y\right)\right]\\ =5x\left(x-4y\right)\\ d,=\left[3\left(x-y\right)-2\left(x+y\right)\right]\left[3\left(x-y\right)+2\left(x+y\right)\right]\\ =\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)\\ =\left(x-5y\right)\left(5x-y\right)\\ f,=\left(x+3\right)\left(x^2-3x+9\right)\\ g,=\left(3x-0,1\right)\left(9x^2+0,3x+0,01\right)\\ h,=\left(5x-1\right)\left(25x^2+5x+1\right)\)

\(a)x^4-y^4=(x^2-y^2)(x^2+y^2)=(x-y)(x+y)(x^2+y^2)\\ b)x^2-3y^2=\\ c)(3x-2y)^2-4(x+y)^2=(3x-2y)^2-[2(x+y)]^2\\=(3x-2y+2x+2y)(3x-2y-2x-2y)=5x(x-4y)\\ d)9(x-y)^2-4(x+y)^2=[3(x-y)]^2-[2(x+y)]^2=(3x-3y+2x+2y)(3x-3y-2x-2y)\\=(5x-y)(x-5y)\\ f)x^3+27=(x+3)(x^2-3x+9)\\ g)27x^3-0,001=(3x-0,1)(9x+0,3x+0,01)\\ h)125x^3-1=(5x-1)(25x^2+5x+1)\)

Lời giải:
a.

$x^2-7x+6=(x^2-x)-(6x-6)=x(x-1)-6(x-1)=(x-1)(x-6)$

b.

$x-3\sqrt{3}x-12\sqrt{3}$ không phân tích được thành nhân tử

c.

$x^2+4x-2$ không phân tích được thành nhân tử với các hệ số nguyên.

a) Ta có: \(x-2y+x^2-4y^2\)

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

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

b) Ta có: \(x^2+2xy+y^2-4x^2y^2\)

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

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

c) Ta có: \(x^6-x^4+2x^3+2x^2\)

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

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

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

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

d) Ta có: \(x^3+3x^2+3x+1-8y^3\)

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

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

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

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

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

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

c, \(x^6-x^4+2x^3+2x^2=x^6+2x^3+1-x^4+2x^2-1\)

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

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

d, \(x^3+3x^2+3x+1-8y^3=\left(x+1\right)^3-\left(2y\right)^3=\left(x+1-2y\right)\left(x+1+2y\right)\)

a) Ta có: \(x-2y+x^2-4y^2\)

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

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

b: Ta có: \(x^2-4x^2y^2+y^2+2xy\)

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

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

a: \(=x\left(x-3\right)-4y\left(x-3\right)\)

=(x-3)(x-4y)

d: \(=\left(x-2\right)\left(x+2\right)+\left(x+2\right)^2\)

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

=2x(x+2)

\(a,=x\left(x-3\right)-4y\left(x-3\right)=\left(x-4y\right)\left(x-3\right)\\ b,=\left(x-1\right)\left(x^2+x+1\right)-4x\left(x-1\right)=\left(x-1\right)\left(x^2-3x+1\right)\\ c,=\left(x-y\right)\left(1-a\right)\\ d,=\left(x-2\right)\left(x-2+x+2\right)=2x\left(x-2\right)\\ e,=x^2\left(x+y\right)-xz\left(x+y\right)=x\left(x-z\right)\left(x+y\right)\\ f,=\left(x-y-2\right)\left(x+y\right)\)

a.

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

\(=\left(x+1+3z\right)\left[\left(x+1\right)^2+3z\left(x+1\right)+9z^2\right]\)

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

b.

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

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

c.

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

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

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

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

a) Ta có: \(x^3+3x^2+3x+1-27z^3\)

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

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

b) Ta có: \(x^2-2xy+y^2-zx+yz\)

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

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

c) Ta có: \(x^4+4x^2-5\)

\(=x^4+4x^2+4-9\)

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

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

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

`a)7x^3y^2+14x^2y^3+7xy^4`

`=7xy^2(x^2+2xy+y^2)`

`=7xy^2(x+y)^2`

______________________________________________

`b)x^2-xy+5x-5y`

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

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

______________________________________________

`c)3x^2-6xy-12+3y^2`

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

`=3[(x-y)^2-4]`

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

a)7x³y²+14x²y³+7xy⁴

=7xy²(x²+2xy+y²)

=7xy²(x+y)²

b)x²-xy + 5x - 5y

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

=(x-y) (x+5)

p

a, \(=\left(xy+1+x-y\right)\left(xy+1-x+y\right)\)

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

\(=2y[x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2]\)

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

c,\(=3\left(x+1\right)^2\left(x^2-x+1\right)y^2\)

câu a, b áp dụng hằng đẳng thức rồi làm nha 

c) 3x⁴y² + 3x³y² + 3xy² + 3y²

= ( 3x⁴y² + 3x³y² ) + ( 3xy² + 3y² )

= 3x³y² ( x + 1) + 3y² ( x + 1 )

= ( 3x³y² + 3y² ) ( x + 1 )

= 3y² ( x³ + 1 ) ( x + 1 )

= 3y² ( x + 1 ) ( x² - x + 1 ) ( x + 1 )

= 3y² ( x + 1 )² ( x² - x + 1 )

a) (xy +1)²- (x-y)²

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

b) (x + y)³ - (x - y)³

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

= 2y(x²+2xy+y²+x²+xy-xy-y²+x²-2xy+y²)

=2y(3x²+y²)

c) 3x⁴y² + 3x³y² + 3xy² + 3y²

=3y²(x⁴+x³+x+1)

= 3y²(x³(x+1)+(x+1)

= 3y²(x+1)(x³+1)

ko bt đúng ko

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

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

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

\(x^2-7x+6=x^2-x-6x+6=x\left(x-1\right)-6\left(x-1\right)=\left(x-1\right)\left(x-6\right)\)