Đáp án:
$\begin{array}{l}
a)Dkxd:a \ge 0;a \ne \dfrac{1}{2}\\
b)P = \left( {\dfrac{{\sqrt {2a} }}{{\sqrt {2a} + 1}} - \dfrac{{\sqrt {2a} }}{{\sqrt {2a} - 1}}} \right):\dfrac{4}{{\sqrt {2a} + 1}}\\
= \dfrac{{\sqrt {2a} \left( {\sqrt {2a} - 1} \right) - \sqrt {2a} \left( {\sqrt {2a} + 1} \right)}}{{\left( {\sqrt {2a} - 1} \right)\left( {\sqrt {2a} + 1} \right)}}.\dfrac{{\left( {\sqrt {2a} + 1} \right)}}{4}\\
= \dfrac{{2a - \sqrt {2a} - 2a - \sqrt {2a} }}{{\sqrt {2a} - 1}}.\dfrac{1}{4}\\
= \dfrac{{ - 2\sqrt {2a} }}{{\sqrt {2a} - 1}}.\dfrac{1}{4}\\
= \dfrac{{\sqrt {2a} }}{{2\left( {1 - \sqrt {2a} } \right)}}\\
c)P = 2\\
\Rightarrow \dfrac{{\sqrt {2a} }}{{2\left( {1 - \sqrt {2a} } \right)}} = 2\\
\Rightarrow \sqrt {2a} = 4\left( {1 - \sqrt {2a} } \right)\\
\Rightarrow \sqrt {2a} = 4 - 4\sqrt {2a} \\
\Rightarrow 5\sqrt {2a} = 4\\
\Rightarrow \sqrt {2a} = \dfrac{4}{5}\\
\Rightarrow 2a = \dfrac{{16}}{{25}}\\
\Rightarrow a = \dfrac{8}{{25}}\left( {tmdk} \right)\\
\text{Vậy}\,a = \dfrac{8}{{25}}
\end{array}$