Đáp án: m<-9/5
Giải thích các bước giải:
$\begin{array}{l}
\left\{ \begin{array}{l}
mx - 2y = 3\\
3x + my = 5
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
3mx - 6y = 9\\
3mx + {m^2}y = 5m
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left( {3mx + {m^2}y} \right) - \left( {3mx - 6y} \right) = 9 + 5m\\
3x + my = 5
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
6y + {m^2}y = 9 + 5m\\
3x + my = 5
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\left( {{m^2} + 6} \right)y = 9 + 5m\\
3x = 5 - my
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
y = \frac{{9 + 5m}}{{{m^2} + 6}}\\
x = \frac{{5 - my}}{3} = \frac{{5 - m.\frac{{9 + 5m}}{{{m^2} + 6}}}}{3}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
y = \frac{{9 + 5m}}{{{m^2} + 6}}\\
x = \frac{{30 - 9m}}{{3\left( {{m^2} + 6} \right)}} = \frac{{10 - 3m}}{{{m^2} + 6}}
\end{array} \right.\\
Khi:\left\{ \begin{array}{l}
x > 0\\
y < 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\frac{{9 + 5m}}{{{m^2} + 6}} < 0\\
\frac{{10 - 3m}}{{{m^2} + 6}} > 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
5m + 9 < 0\\
10 - 3m > 0
\end{array} \right.\left( {do:{m^2} + 6 > 0\forall m} \right)\\
\Rightarrow \left\{ \begin{array}{l}
m < - \frac{9}{5}\\
m < \frac{{10}}{3}
\end{array} \right. \Rightarrow m < - \frac{9}{5}
\end{array}$
