\(x^4+4\)

= \(\left(x^2+2\right)^2-4x^2\)

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

Đa thức này không phân tích được thành nhân tử bạn nhé. 

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

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

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

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

x⁴ - 2x³-2x² -2x -3

=(x⁴+x³)-(3x³+3x²)+(x²+x)-(3x+3)

=x³(x+1)-3x²(x+1)+x(x+1)-3(x+1)

= (x³-3x²+x-3)(x+1)

= ((x³-3x²)+(x-3))(x+1)

= (x²(x-3)+(x-3))(x+1)

=(x²+1)(x-3)(x+1)

\(A=x^2+4=\left(x^2+4x+4\right)-4x=\left(x+2\right)^2-\sqrt{4x}=\left(x+2-\sqrt{4x}\right)\left(x+2+\sqrt{4x}\right)\)

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

a) \(3x^2-6xy+3y^2-12x^2=3\left(x^2-2xy+y^2\right)-12x^2=3\left(x-y\right)^2-12x^2=3\left[\left(x-y\right)^2-4x^2\right]=3\left(x-y-2x\right)\left(x-y+2x\right)=3\left(-x-y\right)\left(3x-y\right)\)

b)\(3x^2y^2-6x^2y^3+12x^2y^2=3x^2y^2\left(1-2y+4\right)=3x^2y^2\left(5-2y\right)\)

c) \(3x^2-3y^2+12x-12y=3\left(x^2-y^2\right)+12\left(x-y\right)=3\left(x-y\right)\left(x+y+4\right)\)

a: \(3x^2-6xy+3y^2-12x^2\)

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

\(=3\left[\left(x-y\right)^2-4x^2\right]\)

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

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

b: \(3x^2y^2-6x^2y^3+12x^2y^2\)

\(=3x^2y^2\left(1-2y+4\right)\)

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

c: Ta có: \(3x^2-3y^2+12x-12y\)

\(=3\left(x-y\right)\left(x+y\right)+12\left(x-y\right)\)

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

a, 2xy^2 ( x^3 -3xy - 4 )

b, x^2 - 4x - 4x +16

= x(x-4) - 4(x-4)

= (x-4) (x-4)

Lời giải:

a.

$2x^4y^2-6x^2y^3-8xy^2=2xy^2(x^3-3xy-4)$

b. 

$x^2-8x+16=x^2-2.4.x+4^2=(x-4)^2$

c.

$12x^2-12=12(x^2-1)=12(x-1)(x+1)$

d. 

$5x^2y-20xy+20y=5y(x^2-4x+4)=5y(x-2)^2$
e.

$3x^2y^2-27y^2=3y^2(x^2-9)=3y^2(x-3)(x+3)$

f. 

$8x^3-27y^3=(2x)^3-(3y)^3=(2x-3y)(4x^2+6xy+9y^2)$

g.

$4x^4-8x^3+4x^2=(2x^2)^2-2.2x^2.2x+(2x)^2$

$=(2x^2-2x)^2=[2x(x-1)]^2=4x^2(x-1)^2$
h.

$7x^2y^2-28y^4=7y^2(x^2-4y^2)=7y^2(x-2y)(x+2y)$

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