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

= (x³+3x²+3x+1)-(4y)³

=(x+1)³-(4y)³

=(x+1-4y)[(x+1)²+(x+1).4y+16y² ]

=(x+1-4y)[(x²+2x+1)+(4xy+4y)+16y²]

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

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

3(x³-y²)

Ta có:x(3x²+4x-7)=x[(3x²-3x)+(7x-7)]=x[3x(x-1)+7(x-1)]=x(x-1)(3x+7)

\(3x^2-6x+3=0\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)

2(x+3)-x³-3x

\(=-x^3-3x+2x+6\)

\(=-x^3-x+6\)

Đa thức này ko phân tích được nha bạn

\(x^3+3x^2+3x+1-27x^3=-26x^3+3x^2+3x+1\)

\(=-26x^3+13x^2-10x^2+5x-2x+1\)

\(=\left(2x-1\right).\left(-13x^2-5x-1\right)\)

\(x^3+3x^2+3x+2\)

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

\(x^3+3x^2+3x+2\)

\(=x^3+2x^2+x^2+2x+x+2\)

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

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

\(x^3+3x^2+3x+2\)

\(=x^3+2x^2+x^2+2x+x+2\)

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

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

\(x^3+3x^2 +3x+2\)

\(=x^3+x^2+x+2x^2+2x+2\)

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

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