Đáp án:
$\begin{array}{l}
1)a)Dkxd:x \ne 0;x \ne - 2;x \ne 2\\
A = \left( {\dfrac{{{x^2}}}{{{x^3} - 4x}} + \dfrac{6}{{6 - 3x}} + \dfrac{1}{{x + 2}}} \right):\\
\left( {x - 2 + \dfrac{{10 - {x^2}}}{{x + 2}}} \right)\\
= \left( {\dfrac{x}{{{x^2} - 4}} - \dfrac{2}{{x - 2}} + \dfrac{1}{{x + 2}}} \right):\\
\dfrac{{\left( {x - 2} \right)\left( {x + 2} \right) + 10 - {x^2}}}{{x + 2}}\\
= \dfrac{{x - 2\left( {x + 2} \right) + x - 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}.\dfrac{{x + 2}}{{{x^2} - 4 + 10 - {x^2}}}\\
= \dfrac{{x - 2x - 4 + x - 2}}{{x - 2}}.\dfrac{1}{6}\\
= \dfrac{{ - 6}}{{x - 2}}.\dfrac{1}{6}\\
= \dfrac{1}{{2 - x}}\\
b)x \ne 0;x \ne 2;x \ne - 2\\
A = 2\\
\Rightarrow \dfrac{1}{{2 - x}} = 2\\
\Rightarrow 2 - x = \dfrac{1}{2}\\
\Rightarrow x = \dfrac{3}{2}\left( {tmdk} \right)\\
c)A < 0\\
\Rightarrow \dfrac{1}{{2 - x}} < 0\\
\Rightarrow 2 - x < 0\\
\Rightarrow x > 2\\
Vậy\,x > 2\\
d)A = \dfrac{1}{{2 - x}} \in Z\\
\Rightarrow \left( {2 - x} \right) \in \left\{ { - 1;1} \right\}\\
\Rightarrow x \in \left\{ {3;1} \right\}\left( {tmdk} \right)\\
2)a)3{x^2} - 7x + 2\\
= 3{x^2} - 6x - x + 2\\
= \left( {x - 2} \right)\left( {3x - 1} \right)
\end{array}$
