Giải thích các bước giải:
\(\begin{array}{l}
a,\\
\left\{ \begin{array}{l}
x + my = 2\\
mx - 2y = 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
mx + {m^2}y = 2m\\
mx - 2y = 1
\end{array} \right.\\
\Rightarrow \left( {mx + {m^2}y} \right) - \left( {mx - 2y} \right) = 2m - 1\\
\Leftrightarrow {m^2}y + 2y = 2m - 1\\
\Leftrightarrow y = \frac{{2m - 1}}{{{m^2} + 2}}\\
x + my = 2\\
\Leftrightarrow x + m.\frac{{2m - 1}}{{{m^2} + 2}} = 2\\
\Leftrightarrow x = 2 - \frac{{2{m^2} - m}}{{{m^2} + 2}} = \frac{{2{m^2} + 4 - 2{m^2} + m}}{{{m^2} + 2}} = \frac{{m + 4}}{{{m^2} + 2}}\\
b,\\
\left\{ \begin{array}{l}
\left( {m - 1} \right)x - my = 3m - 1\\
2x - y = m + 5
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left( {m - 1} \right)x - my = 3m - 1\\
2mx - my = {m^2} + 5m
\end{array} \right.\\
\Rightarrow \left[ {\left( {m - 1} \right)x - my} \right] - \left( {2mx - my} \right) = 3m - 1 - {m^2} - 5m\\
\Leftrightarrow \left( {m - 1} \right)x - 2mx = - {m^2} - 2m - 1\\
\Leftrightarrow \left( {m + 1} \right)x = {\left( {m + 1} \right)^2}\\
\Leftrightarrow x = m + 1\\
\left( {m - 1} \right)x - my = 3m - 1\\
\Leftrightarrow \left( {m - 1} \right)\left( {m + 1} \right) - my = 3m - 1\\
\Leftrightarrow {m^2} - 1 - my = 3m - 1\\
\Leftrightarrow my = {m^2} - 1 - 3m + 1\\
\Leftrightarrow my = {m^2} - 3m\\
\Leftrightarrow y = m - 3
\end{array}\)
