Đáp án:
\[m = 4\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\left\{ \begin{array}{l}
\left( {m - 1} \right)x + y = 3m - 4\\
x + \left( {m - 1} \right)y = m
\end{array} \right.\,\,\,\,\,\left( * \right)\\
TH1:\,\,m = 1\\
\left( * \right) \Leftrightarrow \left\{ \begin{array}{l}
y = - 1\\
x = 1
\end{array} \right. \Rightarrow x + y = 0\,\,\left( L \right)\\
TH2:\,\,m \ne 1\\
\left( * \right) \Leftrightarrow \left\{ \begin{array}{l}
\left( {m - 1} \right)x + y = 3m - 4\\
\left( {m - 1} \right)x + {\left( {m - 1} \right)^2}y = m\left( {m - 1} \right)
\end{array} \right.\\
\Rightarrow \left[ {\left( {m - 1} \right)x + {{\left( {m - 1} \right)}^2}y} \right] - \left[ {\left( {m - 1} \right)x + y} \right] = m\left( {m - 1} \right) - \left( {3m - 4} \right)\\
\Leftrightarrow \left( {{m^2} - 2m} \right)y = {m^2} - 4m + 4\\
\Leftrightarrow m\left( {m - 2} \right)y = {\left( {m - 2} \right)^2}\\
+ )\,\,\,m = 2 \Rightarrow x + y = 2\,\,\,\,\left( L \right)\\
+ )\,\,\,m = 0 \Rightarrow hptvn\\
+ )\,\,\,\left\{ \begin{array}{l}
m \ne 0\\
m \ne 2
\end{array} \right. \Rightarrow y = \frac{{m - 2}}{m}\\
x + \left( {m - 1} \right)y = m\\
\Leftrightarrow x + \frac{{\left( {m - 1} \right)\left( {m - 2} \right)}}{m} = m\\
\Rightarrow x = m - \frac{{{m^2} - 3m + 2}}{m} = \frac{{3m - 2}}{m}\\
x + y = 3\\
\Leftrightarrow \frac{{3m - 2}}{m} + \frac{{m - 2}}{m} = 3\\
\Leftrightarrow 4m - 4 = 3m\\
\Leftrightarrow m = 4\left( {t/m} \right)
\end{array}\)
