giải và biện luận phương trình (3m^2+m-4)x=2m^2+m-3

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