Đáp án: $ - 2 < a < - \frac{1}{3}$
Giải thích các bước giải:
$\begin{array}{l}
\left\{ \begin{array}{l}
x - \left( {a + 3} \right)y = 0\\
\left( {a - 2} \right)x + 4y = a - 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left( {a - 2} \right)x - \left( {a - 2} \right)\left( {a + 3} \right)y = 0\\
\left( {a - 2} \right)x + 4y = a - 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
4y + \left( {{a^2} + a - 6} \right)y = a - 1\\
x - \left( {a + 3} \right)y = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left( {{a^2} + a - 2} \right)y = a - 1\\
x = \left( {a + 3} \right)y
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left( {a - 1} \right)\left( {a + 2} \right)y = a - 1\\
x = \left( {a + 3} \right)y
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a \ne 1\\
y = \frac{1}{{a + 2}}\\
x = \frac{{a + 3}}{{a + 2}}
\end{array} \right.\\
2x + 3y > 5\\
\Rightarrow 2.\frac{{a + 3}}{{a + 2}} + 3.\frac{1}{{a + 2}} > 5\\
\Rightarrow \frac{{\left( {2a + 6} \right) + 3 - 5\left( {a + 2} \right)}}{{a + 2}} > 0\\
\Rightarrow \frac{{2a + 6 + 3 - 5a - 10}}{{a + 2}} > 0\\
\Rightarrow \frac{{3a + 1}}{{a + 2}} < 0\\
\Rightarrow - 2 < a < - \frac{1}{3}
\end{array}$
