Đáp án:
+ Khi m=-1 thì hệ pt:
$\begin{array}{l}
\left\{ \begin{array}{l}
x - 2y = 0\\
- 3x + 4y = - 2
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
2x - 4y = 0\\
- 3x + 4y = - 2
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
2x - 3x = 0 - 2\\
x - 2y = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
- x = - 2\\
y = \frac{x}{2}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = 2\\
y = 1
\end{array} \right.\\
Vậy\,khi\,m = - 1\,thì\,\left( {x;y} \right) = \left( {2;1} \right)\\
+ \left\{ \begin{array}{l}
x - \left( {m + 3} \right)y = 0\\
\left( {m - 2} \right)x + 4y = m - 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left( {m - 2} \right)x - \left( {m - 2} \right)\left( {m + 3} \right)y = 0\left( 1 \right)\\
\left( {m - 2} \right)x + 4y = m - 1\left( 2 \right)
\end{array} \right.\\
\left( 2 \right) - \left( 1 \right) \Rightarrow 4y + \left( {{m^2} + m - 6} \right)y = m - 1\\
\Rightarrow \left( {{m^2} + m - 2} \right)y = m - 1\\
\Rightarrow \left( {m - 1} \right)\left( {m + 2} \right)y = m - 1\\
+ Khi:m = 1 \Rightarrow 0.y = 0 \Rightarrow hpt\,đúng\,\forall x;y\\
+ Khi:m = - 2 \Rightarrow 0.y = - 3 \Rightarrow hpt\,vô\,nghiệm\\
+ Khi:m \ne 1;m \ne - 2 \Rightarrow y = \frac{1}{{m + 2}}\\
\Rightarrow x = \left( {m + 3} \right)y = \frac{{m + 3}}{{m + 2}}
\end{array}$
