Giải thích các bước giải:
\(\begin{array}{l}
C1:\\
DK:x \ne - 1;y \ne 0\\
\left\{ \begin{array}{l}
\frac{2}{{x + 1}} + \frac{3}{y} = - 1\\
\frac{2}{{x + 1}} + \frac{5}{y} = 1
\end{array} \right. \to \left\{ \begin{array}{l}
\frac{2}{y} = 2\\
\frac{2}{{x + 1}} + \frac{3}{y} = - 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = 1\left( {TM} \right)\\
x = - \frac{3}{2}\left( {TM} \right)
\end{array} \right.\\
C2:\\
\left\{ \begin{array}{l}
x = 2 + y\\
2m + my + y = 3
\end{array} \right. \to \left\{ \begin{array}{l}
x = 2 + y\\
y = \frac{{3 - 2m}}{{m + 1}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \frac{{2m + 2 + 3 - 2m}}{{m + 1}} = \frac{5}{{m + 1}}\\
y = \frac{{3 - 2m}}{{m + 1}}
\end{array} \right.\\
Do:\left\{ \begin{array}{l}
x > 0\\
y > 0
\end{array} \right. \to \left\{ \begin{array}{l}
\frac{5}{{m + 1}} > 0\\
\frac{{3 - 2m}}{{m + 1}} > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
m + 1 > 0\\
3 - 2m > 0
\end{array} \right. \to - 1 < m < \frac{3}{2}
\end{array}\)
