\(2x^2+2y^2-x^2z+z-y^2z-2\)

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

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

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

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

-[ ((x²)²+(y²)²+(z²)²-2x²y²-2x²z²+2y²z²)-4y²z²]

- ( (x²-y²-z²)²-(2yz)²)

-( x²-y²-z²-2yz )(x²-y²-z²+2yz)

Sai thì bảo mình đừng k sai a -)

kết  quả là -(x^2-y^2-z^2)^2

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

\(=\left(2xz+x^2-y^2+z^2\right)\left(2xz-x^2+y^2-z^2\right)=\left[\left(x+z\right)^2-y^2\right]\left[y^2-\left(x-z\right)^2\right]\)

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

b: \(=\dfrac{12\left(y-z\right)^4+3\left(y-z\right)^5}{6\left(y-z\right)^2}=2\left(y-z\right)^2+\dfrac{1}{2}\left(y-z\right)^3\)

\(=\dfrac{2\left(x-2y+z\right)^3+4\left(x-2y+z\right)^2}{2\left(x-2y+z\right)}=\left(x-2y+z\right)^2+2\left(x-2y+z\right)\)

a: 6x-2y=2(3x-y)

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

Lời giải:
a. Không phân tích được nữa

b. $x^2(x-y)+4(y-x)=x^2(x-y)-4(x-y)=(x-y)(x^2-4)=(x-y)(x-2)(x+2)$
c. $x^3+2x^2y+xy^2-4x=x(x^2+2xy+y^2-4)$

$=x[(x^2+2xy+y^2)-4]=x[(x+y)^2-2^2]=x(x+y-2)(x+y+2)$

ko phân tích dc

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