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