Đáp án:
$\begin{array}{l}
a)Dkxd:x \ne 2;x \ne - 2\\
A = \left( {\dfrac{x}{{{x^2} - 4}} - \dfrac{2}{{2 - x}} + \dfrac{1}{{x + 2}}} \right):\left( {x - 2 + \dfrac{{10 - {x^2}}}{{x + 2}}} \right)\\
= \dfrac{{x + 2\left( {x + 2} \right) + x - 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}:\dfrac{{\left( {x - 2} \right)\left( {x + 2} \right) + 10 - {x^2}}}{{x + 2}}\\
= \dfrac{{x + 2x + 4 + x - 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}.\dfrac{{x + 2}}{{{x^2} - 4 + 10 - {x^2}}}\\
= \dfrac{{4x + 2}}{{x - 2}}.\dfrac{1}{6}\\
= \dfrac{{2x + 1}}{{3\left( {x - 2} \right)}}\\
b)\left| x \right| = \dfrac{1}{2} \Rightarrow \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = - \dfrac{1}{2}
\end{array} \right.\left( {tmdk} \right)\\
+ Khi:x = \dfrac{1}{2} \Rightarrow A = - \dfrac{1}{2}\\
+ Khi:x = - \dfrac{1}{2} \Rightarrow A = 0\\
c)A < 0\\
\Rightarrow \dfrac{{2x + 1}}{{3\left( {x - 2} \right)}} < 0\\
\Rightarrow - \dfrac{1}{2} < x < 2
\end{array}$
