Đáp án:
$\begin{array}{l}
A = \frac{{x + 2}}{{x + 3}} - \frac{5}{{{x^2} + x - 6}} + \frac{1}{{2 - x}}\\
a)Dkxd:\left\{ \begin{array}{l}
x + 3 \ne 0\\
{x^2} + x - 6 \ne 0\\
2 - x \ne 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ne - 3\\
x \ne 2
\end{array} \right.\\
b)A = \frac{{x + 2}}{{x + 3}} - \frac{5}{{{x^2} + x - 6}} + \frac{1}{{2 - x}}\\
= \frac{{x + 2}}{{x + 3}} - \frac{5}{{\left( {x + 3} \right)\left( {x - 2} \right)}} - \frac{1}{{x - 2}}\\
= \frac{{\left( {x + 2} \right)\left( {x - 2} \right) - 5 - \left( {x + 3} \right)}}{{\left( {x + 3} \right)\left( {x - 2} \right)}}\\
= \frac{{{x^2} - 4 - 5 - x - 3}}{{\left( {x + 3} \right)\left( {x - 2} \right)}}\\
= \frac{{{x^2} - x - 12}}{{\left( {x + 3} \right)\left( {x - 2} \right)}}\\
= \frac{{\left( {x + 3} \right)\left( {x - 4} \right)}}{{\left( {x + 3} \right)\left( {x - 2} \right)}}\\
= \frac{{x - 4}}{{x - 2}}\\
c)x \ne - 3;x \ne 2\\
A = - \frac{3}{4}\\
\Rightarrow \frac{{x - 4}}{{x - 2}} = - \frac{3}{4}\\
\Rightarrow 4\left( {x - 4} \right) = - 3\left( {x - 2} \right)\\
\Rightarrow 4x - 16 = - 3x + 6\\
\Rightarrow 7x = 22\\
\Rightarrow x = \frac{{22}}{7}\left( {tmdk} \right)
\end{array}$
