Đáp án:
$\begin{array}{l}
a)Dkxd:x \ne - 3;x \ne 2\\
Q = \frac{{x + 2}}{{x + 3}} - \frac{5}{{{x^2} + x - 6}} + \frac{1}{{2 - x}}\\
= \frac{{x + 2}}{{x + 3}} - \frac{5}{{\left( {x + 3} \right)\left( {x - 2} \right)}} - \frac{1}{{x - 2}}\\
= \frac{{\left( {x + 2} \right)\left( {x - 2} \right) - 5 - \left( {x + 3} \right)}}{{\left( {x + 3} \right)\left( {x - 2} \right)}}\\
= \frac{{{x^2} - 4 - 5 - x - 3}}{{\left( {x + 3} \right)\left( {x - 2} \right)}}\\
= \frac{{{x^2} - x - 12}}{{\left( {x + 3} \right)\left( {x - 2} \right)}}\\
= \frac{{\left( {x + 3} \right)\left( {x - 4} \right)}}{{\left( {x + 3} \right)\left( {x - 2} \right)}}\\
= \frac{{x - 4}}{{x - 2}}\\
b)x \ne - 3;x \ne 2\\
Q = - \frac{3}{4}\\
\Rightarrow \frac{{x - 4}}{{x - 2}} = - \frac{3}{4}\\
\Rightarrow 4x - 16 = - 3x + 6\\
\Rightarrow 7x = 22\\
\Rightarrow x = \frac{{22}}{7}\left( {tmdk} \right)\\
c){x^2} - 9 = 0\\
\Rightarrow x = 3\left( {do:x \ne - 3} \right)\\
\Rightarrow Q = \frac{{x - 4}}{{x - 2}} = \frac{{3 - 4}}{{3 - 2}} = - 1
\end{array}$
