$\left\{\begin{array}{l} 2mx + y = 2\\ x + 2my = 4-4m\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} 2mx + y = 2\\ 2mx + 4m^2y = 8m-8m^2\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} 2mx + y = 2\\ 2mx + 4m^2y-2mx - y = 8m-8m^2-2\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} 2mx + y = 2\\ (4m^2-1)y=-2(2m-1)^2\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x=\dfrac{2-y}{2m}\\ (2m-1)(2m+1)y=-2(2m-1)^2\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x=\dfrac{4}{2m+1}\\ y=\dfrac{-2(2m-1)}{2m+1}\left(m\ne\pm\dfrac{1}{2}\right)\end{array} \right.\\\Leftrightarrow \left\{\begin{array}{l} x=\dfrac{4}{2m+1}\\ y=\dfrac{-2(2m+1)+4}{2m+1}\left(m\ne\pm\dfrac{1}{2}\right)\end{array} \right.\\\Leftrightarrow \left\{\begin{array}{l} x=\dfrac{4}{2m+1}\\ y=2+\dfrac{4}{2m+1}\left(m\ne\pm\dfrac{1}{2}\right)\end{array} \right.$
Để $x,y \in \mathbb{Z}$
$\Rightarrow 2m+1 \in Ư(4)$
$\Rightarrow m \in \left\{-1;0;\dfrac{-3}{2};\dfrac{1}{2};\dfrac{-5}{2};\dfrac{3}{2}\right\}$
Kết hợp ĐK $\Rightarrow m \in \left\{-1;0;\dfrac{-3}{2};\dfrac{-5}{2};\dfrac{3}{2}\right\}$
