Đáp án:
$\begin{array}{l}
a)DKxd:{x^3} - x \ne 0\\
\Rightarrow x\left( {{x^2} - 1} \right) \ne 0\\
\Rightarrow \left\{ \begin{array}{l}
x \ne 0\\
x \ne 1\\
x \ne - 1
\end{array} \right.\\
b)A = \frac{{{x^3} + 2{x^2} + x}}{{{x^3} - x}}\\
= \frac{{x\left( {{x^2} + 2x + 1} \right)}}{{x\left( {{x^2} - 1} \right)}}\\
= \frac{{x{{\left( {x + 1} \right)}^2}}}{{x\left( {x + 1} \right)\left( {x - 1} \right)}}\\
= \frac{{x\left( {x + 1} \right)}}{{x\left( {x - 1} \right)}}\\
= \frac{{x + 1}}{{x - 1}}\\
c)Dkxd:x \ne 0;x \ne 1;x \ne - 1\\
A = 2\\
\Rightarrow \frac{{x + 1}}{{x - 1}} = 2\\
\Rightarrow x + 1 = 2x - 2\\
\Rightarrow x = 3\left( {tmdk} \right)\\
d)Dkxd:x \ne 0;x \ne 1;x \ne - 1\\
A = \frac{{x + 1}}{{x - 1}} = \frac{{x - 1 + 2}}{{x - 1}} = 1 + \frac{2}{{x - 1}}\\
\Rightarrow \left( {x - 1} \right) \in Ư\left( 2 \right) = \left\{ { - 2; - 1;1;2} \right\}\\
\Rightarrow x \in \left\{ { - 1;0;2;3} \right\}\\
Do:x \ne - 1\\
\Rightarrow x \in \left\{ {0;2;3} \right\}
\end{array}$
