Giải thích các bước giải:
\(\begin{array}{l}
a)\left\{ \begin{array}{l}
\left( {m - 1} \right)x + y = 1\\
2x + \left( {m + 1} \right)y = - 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
y = 1 - \left( {m - 1} \right)x\\
2x + \left( {m - 1} \right)\left( {1 - \left( {m + 1} \right)} \right)x = - 3
\end{array} \right.\\
\Rightarrow 2x - \left( {{m^2} - 1} \right)x = - 3 - m + 1\\
\Leftrightarrow \left( {{m^2} - 3} \right)x = m + 2\\
TH1:m = \sqrt 3 \Rightarrow 0 = \sqrt 3 + 2\left( {VN} \right)\\
TH2:m = - \sqrt 3 \Rightarrow 0 = - \sqrt 3 + 2\left( {VN} \right)\\
TH3:m \ne \pm \sqrt 3 \Rightarrow x = \dfrac{{m + 2}}{{{m^2} - 3}}\\
\Rightarrow y = 1 - \left( {m - 1} \right).\dfrac{{m + 2}}{{{m^2} - 3}} = \dfrac{{ - m - 1}}{{{m^2} - 3}}\\
b)\left\{ \begin{array}{l}
\left( {m + 1} \right)x - my = 5\\
x + my = {m^2} + 4m
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = {m^2} + 4m - my\\
\left( {m + 1} \right)\left( {{m^2} + 4m - my} \right) - my = 5
\end{array} \right.\\
\Rightarrow - m\left( {m + 2} \right)y = 5 - \left( {{m^2} + 4m} \right)\left( {m + 1} \right)\\
\Leftrightarrow m\left( {m + 2} \right)y = {m^3} + 5{m^2} + 4m - 5\\
\Rightarrow y = \dfrac{{{m^3} + 5{m^2} + 4m - 5}}{{{m^2} + 2m}}\left( {m \ne 0;m \ne - 2} \right)\\
y = m + 3 - \dfrac{{2m + 5}}{{{m^2} + 2m}}\\
y \in Z \Rightarrow \dfrac{{2m + 5}}{{{m^2} + 2m}} \in Z \Rightarrow \dfrac{{2{m^2} + 5m}}{{{m^2} + 2m}} = 2 + \dfrac{1}{{m + 2}} \in Z\\
\Rightarrow m + 2 \in \{ - 1;1\} \\
m + 2 = 1 \Rightarrow m = - 1 \Rightarrow y = 5\left( {tm} \right)\\
m + 2 = - 1 \Rightarrow m = - 3 \Rightarrow y = \dfrac{1}{3}\left( {ktm} \right)
\end{array}\)
