\(\Leftrightarrow\left\{{}\begin{matrix}y=2-2mx\\x+2m\left(2-2mx\right)=4-4m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=2-2mx\\x+4m-4m^2x-4+4m=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=2-2mx\\x\left(1-4m^2\right)=-8m+4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-2mx+2\\x\left(2m-1\right)\left(2m+1\right)=4\left(2m-1\right)\end{matrix}\right.\)
Nếu m=1/2 thì hệ có vô số nghiệm
Nếu m=-1/2 thì hệ vô nghiệm
Nếu m<>1/2; m<>-1/2 thì hệ có nghiệm duy nhất là:
\(\left\{{}\begin{matrix}x=\dfrac{4}{2m+1}\\y=-2m\cdot\dfrac{4}{2m+1}+2=\dfrac{-8m+4m+2}{2m+1}=\dfrac{-4m+2}{2m+1}\end{matrix}\right.\)
Để x,y nguyên thì \(\left\{{}\begin{matrix}2m+1\inƯ\left(4\right)\\-4m-2+4⋮2m+1\end{matrix}\right.\Leftrightarrow2m+1\inƯ\left(4\right)\)
mà m nguyên
nên \(2m+1\in\left\{1;-1\right\}\)
=>\(m\in\left\{0;-1\right\}\)