Question

Hộ e ạ

Answer 1

\(1,\\ a,=x\left(x^2-8x+16\right)=x\left(x-4\right)^2\\ b,=6\left(x-y\right)-\left(x-y\right)^2=\left(x-y\right)\left(6-x+y\right)\\ c,=x^2-x+11x-11=\left(x-1\right)\left(x+11\right)\\ 2,\\ a,\Leftrightarrow x\left(x-6\right)+8\left(x-6\right)=0\\ \Leftrightarrow\left(x+8\right)\left(x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=6\end{matrix}\right.\\ b,\Leftrightarrow\left(3x+2-x+1\right)\left(3x+2+x-1\right)=0\\ \Leftrightarrow\left(2x+3\right)\left(4x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=-\dfrac{1}{4}\end{matrix}\right.\)