Giải thích các bước giải:

ĐKXĐ: \(\left\{ \begin{array}{l}
x \ge 0\\
x \ne 1
\end{array} \right.\)

Ta có:

\(\begin{array}{l}
P = \left( {\dfrac{{2\sqrt x  + 2}}{{x\sqrt x  + x - \sqrt x  - 1}} + \dfrac{1}{{\sqrt x  + 1}}} \right):\left( {1 - \dfrac{{\sqrt x }}{{\sqrt x  + 1}}} \right)\\
 = \left( {\dfrac{{2\left( {\sqrt x  + 1} \right)}}{{x.\left( {\sqrt x  + 1} \right) - \left( {\sqrt x  + 1} \right)}} + \dfrac{1}{{\sqrt x  + 1}}} \right):\left( {1 - \dfrac{{\sqrt x }}{{\sqrt x  + 1}}} \right)\\
 = \left( {\dfrac{{2\left( {\sqrt x  + 1} \right)}}{{\left( {\sqrt x  + 1} \right)\left( {x - 1} \right)}} + \dfrac{1}{{\sqrt x  + 1}}} \right):\dfrac{{\left( {\sqrt x  + 1} \right) - \sqrt x }}{{\sqrt x  + 1}}\\
 = \left( {\dfrac{2}{{x - 1}} + \dfrac{1}{{\sqrt x  + 1}}} \right):\dfrac{1}{{\sqrt x  + 1}}\\
 = \left( {\dfrac{2}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}} + \dfrac{1}{{\sqrt x  + 1}}} \right):\dfrac{1}{{\sqrt x  + 1}}\\
 = \dfrac{{2 + \left( {\sqrt x  - 1} \right)}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)}}.\left( {\sqrt x  + 1} \right)\\
 = \dfrac{{\sqrt x  + 1}}{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)}}.\left( {\sqrt x  + 1} \right)\\
 = \dfrac{{\sqrt x  + 1}}{{\sqrt x  - 1}}\\
b,\\
P > 1\\
 \Leftrightarrow \dfrac{{\sqrt x  + 1}}{{\sqrt x  - 1}} - 1 > 0\\
 \Leftrightarrow \dfrac{{\left( {\sqrt x  + 1} \right) - \left( {\sqrt x  - 1} \right)}}{{\sqrt x  - 1}} > 0\\
 \Leftrightarrow \dfrac{2}{{\sqrt x  - 1}} > 0\\
 \Leftrightarrow \sqrt x  - 1 > 0\\
 \Leftrightarrow x > 1\\
c,\\
P = \dfrac{{\sqrt x  + 1}}{{\sqrt x  - 1}} = \dfrac{{\left( {\sqrt x  - 1} \right) + 2}}{{\sqrt x  - 1}} = 1 + \dfrac{2}{{\sqrt x  - 1}}\\
P \in Z \Leftrightarrow \dfrac{2}{{\sqrt x  - 1}} \in Z \Leftrightarrow \sqrt x  - 1 \in \left\{ { \pm 1; \pm 2} \right\}\\
 \Rightarrow \sqrt x  \in \left\{ { - 1;0;2;3} \right\}\\
\sqrt x  \ge 0 \Rightarrow \sqrt x  \in \left\{ {0;2;3} \right\} \Rightarrow x \in \left\{ {0;4;9} \right\}\\
d,\\
\dfrac{1}{P} = \dfrac{{\sqrt x  - 1}}{{\sqrt x  + 1}} = \dfrac{{\left( {\sqrt x  + 1} \right) - 2}}{{\sqrt x  + 1}} = 1 - \dfrac{2}{{\sqrt x  + 1}}\\
\sqrt x  \ge 0,\,\,\forall x\\
 \Rightarrow \sqrt x  + 1 \ge 1 \Rightarrow \dfrac{2}{{\sqrt x  + 1}} \le \dfrac{2}{1} = 2\\
 \Rightarrow \dfrac{1}{P} = 1 - \dfrac{2}{{\sqrt x  + 1}} \ge 1 - 2 =  - 1\\
 \Rightarrow {P_{\min }} =  - 1 \Leftrightarrow x = 0
\end{array}\)