Đáp án:
$\begin{array}{l}
a)Dkxd:x \ne 1\\
D = \left( {\dfrac{{{x^2} + 2}}{{{x^3} - 1}} + \dfrac{x}{{{x^2} + x + 1}} + \dfrac{1}{{1 - x}}} \right):\dfrac{{x - 1}}{2}\\
= \dfrac{{{x^2} + 2 + x\left( {x - 1} \right) - \left( {{x^2} + x + 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}.\dfrac{2}{{x - 1}}\\
= \dfrac{{{x^2} + 2 + {x^2} - x - {x^2} - x - 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}.\dfrac{2}{{x - 1}}\\
= \dfrac{{{x^2} - 2x + 1}}{{{{\left( {x - 1} \right)}^2}\left( {{x^2} + x + 1} \right)}}.2\\
= \dfrac{{2{{\left( {x - 1} \right)}^2}}}{{{{\left( {x - 1} \right)}^2}\left( {{x^2} + x + 1} \right)}}\\
= \dfrac{2}{{{x^2} + x + 1}}\\
b)D = 2\\
\Rightarrow \dfrac{2}{{{x^2} + x + 1}} = 2\\
\Rightarrow {x^2} + x + 1 = 1\\
\Rightarrow {x^2} + x = 0\\
\Rightarrow x\left( {x + 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 0\left( {tmdk} \right)\\
x = - 1\left( {tmdk} \right)
\end{array} \right.\\
Vậy\,x = 0;x = - 1
\end{array}$
