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

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

-[ ((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

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

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

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

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

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

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

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

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

1.A

2.C

3.B

4.C

a

c

b

c

(x-1)y^2-4(x-1)y

Bài 2: 

a: \(3x^2-3xy=3x\left(x-y\right)\)

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

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

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

ỳtct7ct7c7c7t79tc9

\(a,=6x\left(x-2\right)-7\left(x-2\right)=\left(6x-7\right)\left(x-2\right)\)