Đáp án:
$\begin{array}{l}
a)Dkxd:x\# 0;x\# 2;x\# - 2\\
A = \left( {\dfrac{{{x^2}}}{{{x^3} - 4x}} + \dfrac{6}{{6 - 3x}} + \dfrac{1}{{x - 2}}} \right):\left( {x - 2 + \dfrac{{10 - {x^2}}}{{x + 2}}} \right)\\
= \left( {\dfrac{x}{{{x^2} - 4}} + \dfrac{2}{{2 - x}} + \dfrac{1}{{x - 2}}} \right):\dfrac{{\left( {x - 2} \right)\left( {x + 2} \right) + 10 - {x^2}}}{{x + 2}}\\
= \left( {\dfrac{x}{{\left( {x - 2} \right)\left( {x + 2} \right)}} - \dfrac{1}{{x - 2}}} \right).\dfrac{{x + 2}}{{{x^2} - 4 + 10 - {x^2}}}\\
= \dfrac{{x - x - 2}}{{\left( {x + 2} \right)\left( {x - 2} \right)}}.\dfrac{{x + 2}}{6}\\
= \dfrac{{ - 1}}{{3\left( {x - 2} \right)}}\\
= \dfrac{1}{{6 - 3x}}\\
b)\left| x \right| = \dfrac{1}{2} \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{2}\left( {tm} \right)\\
x = - \dfrac{1}{2}\left( {tm} \right)
\end{array} \right.\\
+ Khi:x = \dfrac{1}{2}\\
\Leftrightarrow A = \dfrac{1}{{6 - 3x}} = \dfrac{2}{9}\\
+ Khi:x = - \dfrac{1}{2}\\
\Leftrightarrow A = \dfrac{1}{{6 - 3x}} = \dfrac{2}{{15}}\\
c)A + 2 = 0\\
\Leftrightarrow \dfrac{1}{{6 - 3x}} + 2 = 0\\
\Leftrightarrow \dfrac{1}{{6 - 3x}} = - 2\\
\Leftrightarrow - 12 + 6x = 1\\
\Leftrightarrow 6x = 13\\
\Leftrightarrow x = \dfrac{{13}}{6}\left( {tm} \right)\\
Vay\,x = \dfrac{{13}}{6}\\
d)A < 0\\
\Leftrightarrow \dfrac{1}{{6 - 3x}} < 0\\
\Leftrightarrow 6 - 3x < 0\\
\Leftrightarrow 3x > 6\\
\Leftrightarrow x > 2\\
Vay\,x > 2\\
e)A = \dfrac{1}{{6 - 3x}} \in Z\\
\Leftrightarrow 6 - 3x \in \left\{ { - 1;1} \right\}\\
\Leftrightarrow 3x \in \left\{ {7;5} \right\}\\
\Leftrightarrow x \in \left\{ {\dfrac{7}{3};\dfrac{5}{3}} \right\}\left( {ktm} \right)
\end{array}$
Vậy ko có x nguyên để A nguyên
