Đáp án:
$\begin{array}{l}
Dkxd:x \ge 0;x\# 1\\
\dfrac{1}{{2 + 2\sqrt x }} + \dfrac{1}{{2 - 2\sqrt x }} - \dfrac{{{x^2} + 2}}{{1 - x}}\\
= \dfrac{{1 - \sqrt x + 1 + \sqrt x - 2.\left( {{x^2} + 2} \right)}}{{2\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}}\\
= \dfrac{{2 - 2{x^2} - 4}}{{2\left( {1 - x} \right)}}\\
= \dfrac{{ - 2 - 2{x^2}}}{{2\left( {1 - x} \right)}}\\
= \dfrac{{{x^2} + 1}}{{x - 1}}
\end{array}$
