Đáp án: $ - 2 < m < - 1$
Giải thích các bước giải:
$\begin{array}{l}
\left\{ \begin{array}{l}
- 2{x^2} + 5x - 3 > 0\\
\left( {1 + {m^2}} \right)x + 3m \le 2 - 3x
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
2x - 5x + 3 < 0\\
\left( {1 + {m^2} + 3} \right)x \le 2 - 3m
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left( {2x - 3} \right)\left( {x - 1} \right) < 0\\
\left( {{m^2} + 4} \right)x \le 2 - 3m
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
1 < x < \dfrac{3}{2}\\
x \le \dfrac{{2 - 3m}}{{{m^2} + 4}}
\end{array} \right.\\
\Rightarrow 1 < \dfrac{{2 - 3m}}{{{m^2} + 4}} \le \dfrac{3}{2}\\
\Rightarrow \left\{ \begin{array}{l}
\dfrac{{2 - 3m}}{{{m^2} + 4}} > 1\\
\dfrac{{2 - 3m}}{{{m^2} + 4}} \le \dfrac{3}{2}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
2 - 3m - {m^2} - 4 > 0\\
2\left( {2 - 3m} \right) - 3\left( {{m^2} + 4} \right) \le 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{m^2} + 3m + 2 < 0\\
3{m^2} + 6m + 8 \ge 0\left( {luôn\,đúng} \right)
\end{array} \right.\\
\Rightarrow \left\{ { - 2 < m < - 1} \right.
\end{array}$
