Đáp án:
\[\left[ \begin{array}{l}
m > \frac{{ - 2}}{3}\\
m < \frac{{ - 5}}{2}
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\left\{ \begin{array}{l}
\left( {m + 1} \right)x - my = 4\\
3x - 5y = m
\end{array} \right.\,\,\,\,\,\left( * \right)\\
TH1:\,\,\,m = 0\\
\left( * \right) \Leftrightarrow \left\{ \begin{array}{l}
x = 4\\
3x - 5y = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 4\\
y = \frac{{12}}{5}
\end{array} \right. \Rightarrow x - y = \frac{8}{5} < 2\,\,\,\,\,\left( {t/m} \right)\\
TH2:\,\,\,m \ne 0\\
\left( * \right) \Leftrightarrow \left\{ \begin{array}{l}
5\left( {m + 1} \right)x - 5my = 20\\
3mx - 5my = {m^2}
\end{array} \right.\\
\Leftrightarrow \left[ {5\left( {m + 1} \right)x - 5my} \right] - \left[ {3mx - 5my} \right] = 20 - {m^2}\\
\Leftrightarrow \left( {2m + 5} \right)x = 20 - {m^2}\\
+ )\,\,\,m = - \frac{5}{2} \Rightarrow 0x = \frac{{55}}{4}\,\,\,\left( L \right)\\
+ )\,\,\,m \ne - \frac{5}{2} \Rightarrow x = \frac{{20 - {m^2}}}{{2m + 5}}\\
3x - 5y = m \Leftrightarrow y = \frac{{3x - m}}{5} = \frac{{\frac{{60 - 3{m^2}}}{{2m + 5}} - m}}{5} = \frac{{60 - 3{m^2} - m\left( {2m + 5} \right)}}{{5.\left( {2m + 5} \right)}} = \frac{{60 - 5m - 5{m^2}}}{{5\left( {2m + 5} \right)}} = \frac{{12 - m - {m^2}}}{{2m + 5}}\\
x - y < 2\\
\Leftrightarrow \frac{{20 - {m^2}}}{{2m + 5}} - \frac{{12 - m - {m^2}}}{{2m + 5}} < 2\\
\Leftrightarrow \frac{{8 + m}}{{2m + 5}} - 2 < 0\\
\Leftrightarrow \frac{{8 + m - 4m - 10}}{{2m + 5}} < 0\\
\Leftrightarrow \frac{{3m + 2}}{{2m + 5}} > 0\\
\Leftrightarrow \left[ \begin{array}{l}
m > \frac{{ - 2}}{3}\\
m < \frac{{ - 5}}{2}
\end{array} \right.
\end{array}\)
