\(1,\)
\(a,x^3-2x^2y+x-xz^2=x\left(x^2-2xy+1-z^2\right)\)
\(=x\left[\left(x-1\right)^2-z^2\right]=x\left(x-1-z\right)\left(x-1+z\right)\)
\(b,x^2y+xy^2-x-y=x\left(xy-1\right)+y\left(xy-1\right)\)
\(=\left(x+y\right)\left(xy-1\right)\)
\(2,\)
\(a,\left(x-2\right)\left(x^2+2x+7\right)+2\left(x^2-4\right)-5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+7\right)+2\left(x-2\right)\left(x+2\right)-5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x^2+2x+7+2\left(x+2\right)-5\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+7+2x+4-5\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+4x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x^2+4x+6=0\end{matrix}\right.\)
Còn lại tự túc nhé
\(b,\left(x-2\right)^2=\left(3x-2\right)\left(x-2\right)\)
\(\Leftrightarrow x-2=3x-2\)
\(\Leftrightarrow2x=0\)
\(\Leftrightarrow x=0\)
