Đáp án:
$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
x + 3 \ne 0\\
\left( {x - 2} \right)\left( {x + 3} \right) \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne - 3\\
x \ne 2
\end{array} \right.\\
A = \dfrac{{x + 2}}{{x + 3}} - \dfrac{5}{{\left( {x - 2} \right)\left( {x + 3} \right)}}\\
= \dfrac{{\left( {x + 2} \right)\left( {x - 2} \right) - 5}}{{\left( {x - 2} \right)\left( {x + 3} \right)}}\\
= \dfrac{{{x^2} - 4 - 5}}{{\left( {x - 2} \right)\left( {x + 3} \right)}}\\
= \dfrac{{{x^2} - 9}}{{\left( {x - 2} \right)\left( {x + 3} \right)}}\\
= \dfrac{{\left( {x - 3} \right)\left( {x + 3} \right)}}{{\left( {x - 2} \right)\left( {x + 3} \right)}}\\
= \dfrac{{x - 3}}{{x - 2}}\\
b)x = - 2\left( {tmdk} \right)\\
\Leftrightarrow A = \dfrac{{x - 3}}{{x - 2}} = \dfrac{{ - 2 - 3}}{{ - 2 - 2}} = \dfrac{5}{4}\\
c)A = 0\\
\Leftrightarrow \dfrac{{x - 3}}{{x - 2}} = 0\\
\Leftrightarrow x = 3\left( {tm} \right)\\
Vậy\,x = 3\\
A = 5\\
\Leftrightarrow \dfrac{{x - 3}}{{x - 2}} = 5\\
\Leftrightarrow x - 3 = 5x - 10\\
\Leftrightarrow 4x = 7\\
\Leftrightarrow x = \dfrac{7}{4}\left( {tm} \right)\\
Vậy\,x = \dfrac{7}{4}\\
d)A = \dfrac{{x - 3}}{{x - 2}} = \dfrac{{x - 2 - 1}}{{x - 2}}\\
= 1 - \dfrac{1}{{x - 2}}\\
A \in Z\\
\Leftrightarrow \left( {x - 2} \right) \in \left\{ { - 1;1} \right\}\\
\Leftrightarrow x \in \left\{ {1;3} \right\}\left( {tmdk} \right)\\
Vậy\,x \in \left\{ {1;3} \right\}
\end{array}$
