Đáp án:
\(\left[ \begin{array}{l}
m = 5\\
m = 1\\
m = 4\\
m = 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
mx + 3y = m + 4\\
3x + my = 2m + 5
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{m^2}x + 3my = {m^2} + 4m\\
9x + 3my = 6m + 15
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( {{m^2} - 9} \right)x = {m^2} - 2m - 15\\
y = \dfrac{{m + 4 - mx}}{3}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{\left( {m - 5} \right)\left( {m + 3} \right)}}{{\left( {m - 3} \right)\left( {m + 3} \right)}}\\
y = \dfrac{{m + 4 - mx}}{3}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{m - 5}}{{m - 3}}\\
y = \dfrac{{m + 4 - m.\dfrac{{m - 5}}{{m - 3}}}}{3} = \dfrac{{{m^2} - 3m + 4m - 12 - {m^2} + 5m}}{3}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{m - 5}}{{m - 3}} = \dfrac{{m - 3 - 2}}{{m - 3}} = 1 - \dfrac{2}{{m - 3}}\\
y = \dfrac{{6m - 12}}{3} = 2m - 4
\end{array} \right.\\
DK:\left( {m - 3} \right)\left( {m + 3} \right) \ne 0 \to m \ne \pm 3\\
Do:x \in Z;y \in Z\\
\Leftrightarrow \dfrac{2}{{m - 3}} \in Z\\
\to m - 3 \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
m - 3 = 2\\
m - 3 = - 2\\
m - 3 = 1\\
m - 3 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
m = 5\\
m = 1\\
m = 4\\
m = 2
\end{array} \right.
\end{array}\)
