Đáp án:

\(\begin{array}{l}
a)\dfrac{{2x + 1}}{{x + 3}}\\
b)\dfrac{3}{4}\\
c)\left[ \begin{array}{l}
x >  - \dfrac{1}{2}\\
x <  - 3
\end{array} \right.
\end{array}\)

d) \(x =  - 2\)

Giải thích các bước giải:

\(\begin{array}{l}
a)DK:x \ne \left\{ { - 3; - \dfrac{1}{2};3} \right\}\\
A = \left( {\dfrac{{x - 1}}{{x + 3}} + \dfrac{2}{{x - 3}} + \dfrac{{{x^2} + 3}}{{9 - {x^2}}}} \right):\dfrac{{ - 2}}{{2x + 1}}\\
 = \dfrac{{\left( {x - 1} \right)\left( {x - 3} \right) + 2\left( {x + 3} \right) - {x^2} - 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}.\dfrac{{2x + 1}}{{ - 2}}\\
 = \dfrac{{{x^2} - 4x + 3 + 2x + 6 - {x^2} - 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}.\dfrac{{2x + 1}}{{ - 2}}\\
 = \dfrac{{ - 2x + 6}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}.\dfrac{{2x + 1}}{{ - 2}}\\
 = \dfrac{{2\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}.\dfrac{{2x + 1}}{2}\\
 = \dfrac{{2x + 1}}{{x + 3}}\\
b)\left| {x - 2} \right| = 1\\
 \to \left[ \begin{array}{l}
x - 2 = 1\\
x - 2 =  - 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 3\left( l \right)\\
x = 1
\end{array} \right.\\
Thay:x = 1\\
 \to A = \dfrac{{2.1 + 1}}{{1 + 3}} = \dfrac{3}{4}\\
c)A > 0\\
 \to \dfrac{{2x + 1}}{{x + 3}} > 0\\
 \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
2x + 1 > 0\\
x + 3 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
2x + 1 < 0\\
x + 3 < 0
\end{array} \right.
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x >  - \dfrac{1}{2}\\
x <  - 3
\end{array} \right.
\end{array}\)

\(\begin{array}{l}
d)A = \dfrac{{2x + 1}}{{x + 3}} = \dfrac{{2\left( {x + 3} \right) - 5}}{{x + 3}}\\
 = 2 - \dfrac{5}{{x + 3}}\\
A \in Z \to \dfrac{5}{{x + 3}} \in Z\\
 \to x + 3 \in U\left( 5 \right)\\
 \to \left[ \begin{array}{l}
x + 3 = 5\\
x + 3 =  - 5\\
x + 3 = 1\\
x + 3 =  - 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 2\\
x =  - 8\\
x =  - 2\\
x =  - 4
\end{array} \right. \to \left[ \begin{array}{l}
A = 1\\
A = 3\\
A =  - 3\\
A = 7
\end{array} \right.\\
 \to x =  - 2\left( {TMDK} \right)\\
KL: x=-2
\end{array}\)