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