Đáp án:
$\begin{array}{l}
\left\{ \begin{array}{l}
mx + y = 2m\\
x + my = m + 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
mx + y = 2m\\
mx + {m^2}y = {m^2} + m
\end{array} \right.\\
\Rightarrow {m^2}y - y = {m^2} + m - 2m\\
\Rightarrow \left( {{m^2} - 1} \right).y = {m^2} - m\\
\Rightarrow \left( {m - 1} \right).\left( {m + 1} \right).y = m\left( {m - 1} \right)\left( * \right)
\end{array}$
+Khi m=1 => 0.y=0
=>Hpt có vô số nghiệm x,y nguyên
$\begin{array}{l}
Khi:m \ne 1;m \ne - 1\\
\Rightarrow y = \dfrac{m}{{m + 1}}\\
\Rightarrow x = m + 1 - m.y\\
= m + 1 - \dfrac{{{m^2}}}{{m + 1}}\\
= \dfrac{{{{\left( {m + 1} \right)}^2} - {m^2}}}{{m + 1}}\\
= \dfrac{{2m + 1}}{{m + 1}}\\
\left\{ {x;y \in Z} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\dfrac{m}{{m + 1}} \in Z\\
\dfrac{{2m + 1}}{{m + 1}} \in Z
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
1 - \dfrac{1}{{m + 1}} \in Z\\
\dfrac{{2m + 2 - 1}}{{m + 1}} = 2 - \dfrac{1}{{m + 1}} \in Z
\end{array} \right.\\
\Rightarrow \dfrac{1}{{m + 1}} \in Z\\
\Rightarrow \left[ \begin{array}{l}
m + 1 = 1\\
m + 1 = - 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
m = 0\left( {tm} \right)\\
m = - 2\left( {tm} \right)
\end{array} \right.
\end{array}$
Vậy m=1; m=0 hoặc m=-2.
