Đáp án:
$\begin{array}{l}
{m^2}\left( {x - 2} \right) + m\left( {x + 3} \right) = 2\left( {3x - 1} \right)\\
\Rightarrow {m^2}.x + mx - 2{m^2} + 3m = 6x - 2\\
\Rightarrow \left( {{m^2} + m - 6} \right).x = 2{m^2} - 3m - 2\\
\Rightarrow \left( {m - 2} \right)\left( {m + 3} \right).x = \left( {2m + 1} \right)\left( {m - 2} \right)\\
+ Khi:m = 2 \Rightarrow 0.x = 0 \Rightarrow \forall x\\
+ Khi:m = - 3 \Rightarrow 0.x = 25\left( {vô\,nghiệm} \right)\\
+ Khi:m \ne - 3;m \ne 2 \Rightarrow x = \frac{{2m + 1}}{{m + 3}}
\end{array}$
