Giải thích các bước giải:
$\begin{array}{l}
+ xet\,pt\,3{m^2} + m - 4 = 0 \Leftrightarrow \left[ \begin{array}{l}
m = 1\\
m = - \frac{4}{3}
\end{array} \right.\\
+ 2{m^2} + m - 3 = 0 \Leftrightarrow \left[ \begin{array}{l}
m = 1\\
m = - \frac{3}{2}
\end{array} \right.\\
- TH1:\left\{ \begin{array}{l}
3{m^2} + m - 4 = 0\\
2{m^2} + m - 3 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
m = 1\,hoac\,m = \frac{{ - 4}}{3}\\
m = 1\,hoac\,m = \frac{{ - 3}}{2}
\end{array} \right. \Leftrightarrow m = 1\\
thi\,pt \Leftrightarrow 0.x = 0\,dung\,\forall x\\
vay\,m = 1\,thi\,pt\,co\,tap\,nghiem\,x = R\\
- TH2:\,\left\{ \begin{array}{l}
3{m^2} + m - 4 \ne 0\\
2{m^2} + m - 3 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
m \ne 1va\,m \ne - \frac{4}{3}\\
m = 1\,hoac\,m = - \frac{3}{2}
\end{array} \right. \Leftrightarrow m = - \frac{3}{2}\\
thi\,pt \Leftrightarrow \left( {3{m^2} + m - 4} \right)x = 0 \Leftrightarrow x = 0\\
vay\,m = - \frac{3}{2}thi\,x = 0\\
- TH3:\left\{ \begin{array}{l}
3{m^2} + m - 4 \ne 0\\
2{m^2} + m - 3 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
m \ne 1;\,m \ne - \frac{4}{3}\\
m \ne 1\,;m \ne - \frac{3}{2}
\end{array} \right. \Leftrightarrow \left\{ {m \ne 1;m \ne - \frac{4}{3};m \ne - \frac{3}{2}} \right.\\
thi\,pt\,co\,nghiem\,x = \frac{{3{m^2} + m - 4}}{{2{m^2} + m - 3}} = \frac{{3m + 4}}{{2m + 3}}
\end{array}$
