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

\(=x^5-2x^4+x^3-x^4+2x^3-x^2\)

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

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

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

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

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

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

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

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

\(x^5-3x^4+3x^3-x^2\)

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

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

hk tốt

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

\(x^4-4x^3-2x^2-3x+2\)

\(\Leftrightarrow x^4+x^3-5x^3+x^2-5x^2+2x^2-5x+2x+2\)

\(\Leftrightarrow x^4+x^3+x^2-5x^3-5x^2-5x+2x^2+2x+2\)

\(\Leftrightarrow x^2\left(x^2+x+1\right)-5x\left(x^2+x+1\right)+2\left(x^2+x+1\right)\)

\(\Leftrightarrow\left(x^2-5x+2\right)\left(x^2+x+1\right)\)

Xin tick ạ !!!

\(x^3-3x^2+6x-4\)

\(=x^3-2x^2+4x-x^2+2x-4\)

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

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

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

x^3 - 3x^2 + 6x - 4 

<=> x^3-3x^2+3x-1+3x-3

<=>(x-1)^3+3(x-1)

<=>(x-1)+((x-1)^2+3)

<=>(x-1)+(x^2-2x+4)

\(x^4+3x^2-2x+3\)

\(=x^4+x^3-x^3+3x^2+x^2-x^2-3x+x+3\)

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

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

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

bạn sử dụng phương pháp hệ số bất định nha để tìm đáp số 

\(a,x^4+5x^3-8x-40=x^3\left(x+5\right)-8\left(x+5\right)\\ =\left(x^3-8\right)\left(x+5\right)=\left(x-2\right)\left(x^2+2x+4\right)\left(x+5\right)\\ b,3x^2-6x-12y^2+3=3\left(x^2-2x-4y^2+1\right)\\ =3\left[\left(x-1\right)^2-4y^2\right]=3\left(x-2y-1\right)\left(x+2y-1\right)\)