Giải thích các bước giải:

a,

Ta có:

\(\begin{array}{l}
P = \left( {\frac{{\sqrt a  + \sqrt b }}{{1 - \sqrt {ab} }} + \frac{{\sqrt a  - \sqrt b }}{{1 + \sqrt {ab} }}} \right):\left( {1 + \frac{{a + b + 2ab}}{{1 - ab}}} \right)\\
 = \frac{{\left( {\sqrt a  + \sqrt b } \right)\left( {1 + \sqrt {ab} } \right) + \left( {\sqrt a  - \sqrt b } \right)\left( {1 - \sqrt {ab} } \right)}}{{\left( {1 - \sqrt {ab} } \right)\left( {1 + \sqrt {ab} } \right)}}:\frac{{1 - ab + a + b + 2ab}}{{1 - ab}}\\
 = \frac{{\sqrt a  + a\sqrt b  + \sqrt b  + b\sqrt a  + \sqrt a  - a\sqrt b  - \sqrt b  + b\sqrt a }}{{\left( {1 - \sqrt {ab} } \right)\left( {1 + \sqrt {ab} } \right)}}:\frac{{a + b + ab + 1}}{{1 - ab}}\\
 = \frac{{2\sqrt a  + 2b\sqrt a }}{{1 - ab}}.\frac{{1 - ab}}{{a\left( {1 + b} \right) + \left( {b + 1} \right)}}\\
 = \frac{{2\sqrt a \left( {b + 1} \right)}}{{\left( {a + 1} \right)\left( {b + 1} \right)}}\\
 = \frac{{2\sqrt a }}{{a + 1}}\\
\end{array}\)

b,

\(\begin{array}{l}
P = \frac{{2\sqrt a }}{{a + 1}} > 0,\,\,\,\forall a > 0\\
{\left( {\sqrt a  - 1} \right)^2} \ge 0,\,\,\,\forall a > 0\\
 \Rightarrow a + 1 \ge 2\sqrt a  \Rightarrow P = \frac{{2\sqrt a }}{{a + 1}} \le \frac{{a + 1}}{{a + 1}} = 1\\
 \Rightarrow 0 < P \le 1 \Rightarrow 0 < 2P \le 2\\
P \in Z \Rightarrow \left[ \begin{array}{l}
2P = 1\\
2P = 2
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
P = \frac{1}{2}\\
P = 1
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\sqrt a  = 2 \pm \sqrt 3 \\
\sqrt a  = 1
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
a = 7 \pm 4\sqrt 3 \\
a = 1
\end{array} \right.
\end{array}\)