Đáp án:
2) 1<m<2
Giải thích các bước giải:
\(\begin{array}{l}
\left\{ \begin{array}{l}
- 2x - 2y = - 6\\
2x + my = m + 5
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( {m - 2} \right)y = m - 1\\
x + y = 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \dfrac{{m - 1}}{{m - 2}}\\
x = 3 - \dfrac{{m - 1}}{{m - 2}} = \dfrac{{3m - 6 - m + 1}}{{m - 2}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = \dfrac{{m - 1}}{{m - 2}}\\
x = \dfrac{{2m - 5}}{{m - 2}}
\end{array} \right.\\
DK:m \ne 2\\
1)x + 2y = 3\\
\to \dfrac{{2m - 5}}{{m - 2}} + 2.\dfrac{{m - 1}}{{m - 2}} = 3\\
\to \dfrac{{2m - 5 + 2m - 2 - 3m + 6}}{{m - 2}} = 0\\
\to m - 1 = 0\\
\to m = 1\\
2)Do:x > 1;y < 0\\
\to \left\{ \begin{array}{l}
\dfrac{{m - 1}}{{m - 2}} < 0\\
\dfrac{{2m - 5}}{{m - 2}} > 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{{m - 1}}{{m - 2}} < 0\\
\dfrac{{2m - 5 - m + 2}}{{m - 2}} > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{{m - 1}}{{m - 2}} < 0\\
\dfrac{{m - 3}}{{m - 2}} > 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m - 2 > 0\\
m - 1 < 0\\
m - 3 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
m - 2 < 0\\
m - 1 > 0\\
m - 3 < 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m > 2\\
m < 1\\
m > 3
\end{array} \right.\left( l \right)\\
\left\{ \begin{array}{l}
m < 2\\
m > 1\\
m < 3
\end{array} \right.
\end{array} \right.\\
\to 1 < m < 2
\end{array}\)
