làm rồi

\(1,=5x\left(1-4x+4x^2\right)=5x\left(2x-1\right)^2\\ 2,=x\left(x-2y\right)-3\left(x-2y\right)=\left(x-3\right)\left(x-2y\right)\\ 3,=4x^2-\left(y+3\right)^2=\left(2x+y+3\right)\left(2x-y-3\right)\)

Chia 2 dư 1 \(\Rightarrow A\) lẻ

Mà chia 5 dư 1 \(\Rightarrow A\) tận cùng \(1,6\)

\(\Rightarrow A\) tận cùng 1

\(\Rightarrow A=\overline{x4691}:9\) dư 1

\(\Rightarrow x+4+6+9+1=x+20:9\) dư 1

\(\Rightarrow x+20=28\Rightarrow x=8\)

Vậy \(A=84691\)

\(\Rightarrow\dfrac{2x}{6}=\dfrac{3y}{6}=\dfrac{z}{6}\Rightarrow\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{z}{6}=\dfrac{x+y+z}{3+2+6}=\dfrac{11}{11}=1\\ \Rightarrow\left\{{}\begin{matrix}x=3\\y=2\\z=6\end{matrix}\right.\Rightarrow z-x=6-3=3\left(A\right)\)

\(Sửa:y=\left(2m-2\right)x-3\\ \text{//}\Leftrightarrow\left\{{}\begin{matrix}2m-2=2\\2k+1\ne-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m=2\\k\ne-2\end{matrix}\right.\)

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