Đáp án: m=0 hoặc m=-1
Giải thích các bước giải:
$\begin{array}{l}
\left\{ \begin{array}{l}
x - \left( {m + 1} \right)y = 1\\
4x - y = - 2
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
4x - 4\left( {m + 1} \right)y = 4\\
4x - y = - 2
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left( {1 - 4m - 4} \right)y = 6\\
x - \left( {m + 1} \right)y = 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left( { - 3 - 4m} \right)y = 6\left( 1 \right)\\
x = 1 + \left( {m + 1} \right)y
\end{array} \right.\\
\left( 1 \right) \Rightarrow \left\{ \begin{array}{l}
- 3 - 4m \ne 0\\
6 \vdots \left( { - 3 - 4m} \right)
\end{array} \right.\\
\Rightarrow \left( { - 3 - 4m} \right) \in {\rm{\{ }} - 3; - 1;1;3{\rm{\} }}\left( {do: - 3 - 4m\,lẻ} \right)\\
\Rightarrow m \in {\rm{\{ }}0; - \frac{1}{2}; - 1; - \frac{3}{2}{\rm{\} }}\\
\Rightarrow m \in {\rm{\{ }}0; - 1\} \\
+ )Khi:m = 0 \Rightarrow \left\{ \begin{array}{l}
y = \frac{6}{{ - 3 - 4m}} = - 2\\
x = 1 + \left( {m + 1} \right)y = - 1
\end{array} \right.\left( {tm} \right)\\
+ )Khi:m = - 1 \Rightarrow \left\{ \begin{array}{l}
y = 6\\
x = 1
\end{array} \right.\left( {tm} \right)
\end{array}$
