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

1.A

2.C

3.B

4.C

a

c

b

c

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

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

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

3) \(xy\left(x+y\right)-x-y=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\) (xem lại đề sửa -2x thành -x mới đúng)

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

\(R=\left(2x\right)^2-2.2x.y+y^2\\ =\left(2x-y\right)^2\\ =>D\)

a) Ta có: \(x^2-2xy+y^2-2x+2y\)

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

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

b) Ta có: \(x^2-4x+4-x^2y+2xy\)

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

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

c) Ta có: \(ax^2-3axy-x^2+6xy-9y^2\)

\(=ax\left(x-3y\right)-\left(x^2-6xy+9y^2\right)\)

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

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

d) Ta có: \(2a^2x-5a^2y-4x^2+30xy-25y^2\)

\(=a^2\left(2x-5y\right)-\left(4x^2-30xy+25y^2\right)\)

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

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

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

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

-[ ((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

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

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

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