Đáp án:
$\begin{array}{l}
a)Dkxd:x\# 2;x\# - 2\\
b)A = \left( {\dfrac{x}{{{x^2} - 4}} + \dfrac{2}{{2 - x}} + \dfrac{1}{{x + 2}}} \right):\left( {x - 2 + \dfrac{{10 - {x^2}}}{{x + 2}}} \right)\\
= \dfrac{{x - 2\left( {x + 2} \right) + x - 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}:\dfrac{{\left( {x - 2} \right)\left( {x + 2} \right) + 10 - {x^2}}}{{x + 2}}\\
= \dfrac{{x - 2x - 4 + x - 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}.\dfrac{{x + 2}}{{{x^2} - 4 + 10 - {x^2}}}\\
= \dfrac{{ - 6}}{{x - 2}}.\dfrac{1}{6}\\
= \dfrac{1}{{2 - x}}\\
c)\left| x \right| = \dfrac{1}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{2}\left( {tm} \right) \Leftrightarrow A = \dfrac{1}{{2 - x}} = \dfrac{2}{3}\\
x = - \dfrac{1}{2}\left( {tm} \right) \Leftrightarrow A = \dfrac{2}{5}
\end{array} \right.\\
Vậy\,A = \dfrac{2}{3};A = \dfrac{2}{5}\\
d)A < 0\\
\Leftrightarrow \dfrac{1}{{2 - x}} < 0\\
\Leftrightarrow 2 - x < 0\\
\Leftrightarrow x > 2\\
Vậy\,x > 2\\
e)A \ge 4\\
\Leftrightarrow \dfrac{1}{{2 - x}} \ge 4\\
\Leftrightarrow \dfrac{1}{{2 - x}} - 4 \ge 0\\
\Leftrightarrow \dfrac{{1 - 4\left( {2 - x} \right)}}{{2 - x}} \ge 0\\
\Leftrightarrow \dfrac{{4x - 7}}{{x - 2}} \le 0\\
\Leftrightarrow \dfrac{7}{4} \le x < 2\\
Vậy\,\dfrac{7}{4} \le x < 2\\
f)A = 3\\
\Leftrightarrow \dfrac{1}{{2 - x}} = 3\\
\Leftrightarrow 2 - x = \dfrac{1}{3}\\
\Leftrightarrow x = \dfrac{5}{3}\left( {tmdk} \right)\\
Vậy\,x = \dfrac{5}{3}
\end{array}$
