Đáp án: $\,\frac{2}{3} < m < \frac{{4 + \sqrt {13} }}{6}$
Giải thích các bước giải:
$\begin{array}{l}
\left( {2 - 3m} \right){x^2} + 3m - x - m < 0\forall x\\
\Leftrightarrow \left( {2 - 3m} \right){x^2} - x + 2m < 0\forall x\\
\Leftrightarrow \left\{ \begin{array}{l}
2 - 3m < 0\\
\Delta < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > \frac{2}{3}\\
1 - 4.\left( {2 - 3m} \right).2m < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > \frac{2}{3}\\
12{m^2} - 16m + 1 < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > \frac{2}{3}\\
\frac{{4 - \sqrt {13} }}{6} < m < \frac{{4 + \sqrt {13} }}{6}
\end{array} \right.\\
Vậy\,\frac{2}{3} < m < \frac{{4 + \sqrt {13} }}{6}
\end{array}$
