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