Đáp án:
$\begin{array}{l}
a)\\
Dkxd:\left\{ \begin{array}{l}
a \ge 0\\
1 - a \ne 0\\
1 - a\sqrt a \ne 0\\
2\sqrt a - 1 \ne 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
a \ge 0\\
a \ne 1\\
a \ne \frac{1}{4}
\end{array} \right.\\
P = 1 + \left( {\frac{{2a + \sqrt a - 1}}{{1 - a}} - \frac{{2a\sqrt a - \sqrt a + a}}{{1 - a\sqrt a }}} \right).\frac{{a - \sqrt a }}{{2\sqrt a - 1}}\\
= 1 + \left[ {\frac{{2a + 2\sqrt a - \sqrt a - 1}}{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a } \right)}} - \frac{{\sqrt a \left( {2a + \sqrt a - 1} \right)}}{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a + a} \right)}}} \right].\frac{{\sqrt a \left( {\sqrt a - 1} \right)}}{{2\sqrt a - 1}}\\
= 1 + \left[ {\frac{{\left( {\sqrt a + 1} \right).\left( {2\sqrt a - 1} \right)}}{{\left( {1 - \sqrt a } \right)\left( {\sqrt a + 1} \right)}} - \frac{{\sqrt a .\left( {\sqrt a + 1} \right).\left( {2\sqrt a - 1} \right)}}{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a + a} \right)}}} \right].\frac{{\sqrt a \left( {\sqrt a - 1} \right)}}{{2\sqrt a - 1}}\\
= 1 + \left[ {\frac{1}{{1 - \sqrt a }} - \frac{{\sqrt a .\left( {\sqrt a + 1} \right)}}{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a + a} \right)}}} \right].\left( {2\sqrt a - 1} \right).\frac{{\sqrt a \left( {\sqrt a - 1} \right)}}{{2\sqrt a - 1}}\\
= 1 + \frac{{1 + \sqrt a + a - a - \sqrt a }}{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a + a} \right)}}.\sqrt a .\left( {\sqrt a - 1} \right)\\
= 1 + \frac{{ - \sqrt a }}{{1 + \sqrt a + a}}\\
= \frac{{a + 1}}{{a + \sqrt a + 1}}\\
b)a \ge 0;a \ne 1;a \ne \frac{1}{4}\\
P = \frac{{\sqrt 6 }}{{1 + \sqrt 6 }}\\
\Rightarrow \frac{{a + 1}}{{a + \sqrt a + 1}} = \frac{{\sqrt 6 }}{{1 + \sqrt 6 }}\\
\Rightarrow \frac{{a + \sqrt a + 1 - \sqrt a }}{{a + \sqrt a + 1}} = \frac{{\sqrt 6 + 1 - 1}}{{\sqrt 6 + 1}}\\
\Rightarrow 1 - \frac{{\sqrt a }}{{a + \sqrt a + 1}} = 1 - \frac{1}{{\sqrt 6 + 1}}\\
\Rightarrow \frac{{\sqrt a }}{{a + \sqrt a + 1}} = \frac{1}{{\sqrt 6 + 1}}\\
\Rightarrow \left( {\sqrt 6 + 1} \right).\sqrt a = a + \sqrt a + 1\\
\Rightarrow a - \sqrt 6 .\sqrt a + 1 = 0\\
\Rightarrow \sqrt a = \frac{{\sqrt 6 \pm \sqrt 2 }}{2}\\
\Rightarrow a = 2 + \sqrt 3 /a = 2 - \sqrt 3 \\
c)P = \frac{{a + 1}}{{a + \sqrt a + 1}} > \frac{2}{3}\\
\Rightarrow 3a + 3 > 2a + 2\sqrt a + 1\\
\Rightarrow a - 2\sqrt a + 2 > 0\\
\Rightarrow a - 2\sqrt a + 1 + 1 > 0\\
\Rightarrow {\left( {\sqrt a - 1} \right)^2} + 1 > 0\,đúng\forall x > 0;x \ne 1;x \ne \frac{1}{4}\\
\Rightarrow P > \frac{2}{3}
\end{array}$
