Đáp án:

$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
x\# 1\\
x\#  - 1\\
x\# 0
\end{array} \right.\\
A = \left( {\dfrac{{x + 1}}{{x - 1}} - \dfrac{{x - 1}}{{x + 1}}} \right):\dfrac{{2x}}{{5x - 5}}\\
 = \dfrac{{{{\left( {x + 1} \right)}^2} - {{\left( {x - 1} \right)}^2}}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}.\dfrac{{5\left( {x - 1} \right)}}{{2x}}\\
 = \dfrac{{{x^2} + 2x + 1 - {x^2} + 2x - 1}}{{x + 1}}.\dfrac{5}{{2x}}\\
 = \dfrac{{4x}}{{x + 1}}.\dfrac{5}{{2x}}\\
 = \dfrac{{10}}{{x + 1}}\\
b)x =  - 3\left( {tmdk} \right)\\
A = \dfrac{{10}}{{ - 3 + 1}} =  - 5\\
c)\left| {x - 2} \right| = 4 - 2x\left( {x \le 2} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 2 = 4 - 2x\\
x - 2 = 2x - 4
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
3x = 6\\
x = 2\left( {tm} \right)
\end{array} \right.\\
 \Leftrightarrow x = 2\left( {tmdk} \right)\\
 \Leftrightarrow A = \dfrac{{10}}{{2 + 1}} = \dfrac{{10}}{3}\\
d)A = 2\\
 \Leftrightarrow \dfrac{{10}}{{x + 1}} = 2\\
 \Leftrightarrow x + 1 = 5\\
 \Leftrightarrow x = 4\left( {tmdk} \right)\\
Vậy\,x = 4\\
e)A < 0\\
 \Leftrightarrow \dfrac{{10}}{{x + 1}} < 0\\
 \Leftrightarrow x + 1 < 0\\
 \Leftrightarrow x <  - 1\\
Vậy\,x <  - 1\\
f)A = \dfrac{{10}}{{x + 1}} \in Z\\
 \Leftrightarrow \left( {x + 1} \right) \in \left\{ { - 10; - 5; - 2; - 1;1;2;5;10} \right\}\\
 \Leftrightarrow x \in \left\{ { - 11; - 6; - 3; - 2;0;1;4;9} \right\}\\
Do:x\# 1;x\#  - 1;x\# 0\\
 \Leftrightarrow x \in \left\{ { - 11; - 6; - 3; - 2;4;9} \right\}\\
g)A >  - 1\\
 \Leftrightarrow \dfrac{{10}}{{x + 1}} >  - 1\\
 \Leftrightarrow \dfrac{{10}}{{x + 1}} + 1 > 0\\
 \Leftrightarrow \dfrac{{10 + x + 1}}{{x + 1}} > 0\\
 \Leftrightarrow \dfrac{{x + 11}}{{x + 1}} > 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 11 > 0\\
x + 1 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 11 < 0\\
x + 1 < 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x >  - 11\\
x >  - 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x <  - 11\\
x <  - 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x >  - 1\\
x <  - 11
\end{array} \right.\\
Vậy\,x <  - 11\,hoặc\,x >  - 1;x\# 0;x\# 1
\end{array}$