Đáp án:
$\begin{array}{l}
a)Dkxd:x \ne 1;x \ne - 1;x \ne - \dfrac{1}{2}\\
A = \left( {\dfrac{1}{{x - 1}} - \dfrac{x}{{1 - {x^3}}}.\dfrac{{{x^2} + x + 1}}{{x + 1}}} \right):\dfrac{{2x + 1}}{{{x^2} + 2x + 1}}\\
= \left( {\dfrac{1}{{x - 1}} - \dfrac{x}{{\left( {1 - x} \right)\left( {{x^2} + x + 1} \right)}}.\dfrac{{{x^2} + x + 1}}{{x + 1}}} \right)\\
.\dfrac{{{{\left( {x + 1} \right)}^2}}}{{2x + 1}}\\
= \left( {\dfrac{1}{{x - 1}} - \dfrac{x}{{\left( {1 - x} \right)\left( {x + 1} \right)}}} \right).\dfrac{{{{\left( {x + 1} \right)}^2}}}{{2x + 1}}\\
= \dfrac{{x + 1 + x}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}.\dfrac{{{{\left( {x + 1} \right)}^2}}}{{2x + 1}}\\
= \dfrac{{x + 1}}{{x - 1}}
\end{array}$
