Đáp án:

1) \(\dfrac{{\sqrt x  + 2}}{{\sqrt x  - 1}}\)

2) \(\left[ \begin{array}{l}
0 \le x < 1\\
4 < x
\end{array} \right.\)

3) \(\left[ \begin{array}{l}
x = 16\\
x = 4\\
x = 0
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
a)P = \dfrac{{x + 1 + \sqrt x }}{{x + 1}}:\left( {\dfrac{1}{{\sqrt x  - 1}} - \dfrac{{2\sqrt x }}{{x\left( {\sqrt x  - 1} \right) + \left( {\sqrt x  - 1} \right)}}} \right) - \dfrac{{\left( {\sqrt x  + 1} \right)\left( {x + \sqrt x  + 1} \right)}}{{x + \sqrt x  + 1}}\\
 = \dfrac{{x + 1 + \sqrt x }}{{x + 1}}:\dfrac{{x + 1 - 2\sqrt x }}{{\left( {\sqrt x  - 1} \right)\left( {x + 1} \right)}} - \left( {\sqrt x  + 1} \right)\\
 = \dfrac{{x + 1 + \sqrt x }}{{x + 1}}.\dfrac{{\left( {\sqrt x  - 1} \right)\left( {x + 1} \right)}}{{{{\left( {\sqrt x  - 1} \right)}^2}}} - \left( {\sqrt x  + 1} \right)\\
 = \dfrac{{x + 1 + \sqrt x }}{{\sqrt x  - 1}} - \left( {\sqrt x  + 1} \right)\\
 = \dfrac{{x + 1 + \sqrt x  - x + 1}}{{\sqrt x  - 1}}\\
 = \dfrac{{\sqrt x  + 2}}{{\sqrt x  - 1}}\\
b)P < 4\\
 \to \dfrac{{\sqrt x  + 2}}{{\sqrt x  - 1}} < 4\\
 \to \dfrac{{\sqrt x  + 2 - 4\sqrt x  + 4}}{{\sqrt x  - 1}} < 0\\
 \to \dfrac{{6 - 3\sqrt x }}{{\sqrt x  - 1}} < 0\\
 \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
6 - 3\sqrt x  > 0\\
\sqrt x  - 1 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
6 - 3\sqrt x  < 0\\
\sqrt x  - 1 > 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
2 > \sqrt x \\
\sqrt x  < 1
\end{array} \right.\\
\left\{ \begin{array}{l}
2 < \sqrt x \\
\sqrt x  > 1
\end{array} \right.
\end{array} \right.\\
 \to \left[ \begin{array}{l}
\sqrt x  < 1\\
2 < \sqrt x 
\end{array} \right. \to \left[ \begin{array}{l}
0 \le x < 1\\
4 < x
\end{array} \right.\\
3)P = \dfrac{{\sqrt x  + 2}}{{\sqrt x  - 1}} = \dfrac{{\sqrt x  - 1 + 3}}{{\sqrt x  - 1}} = 1 + \dfrac{3}{{\sqrt x  - 1}}\\
P \in Z \to \dfrac{3}{{\sqrt x  - 1}} \in Z\\
 \to \sqrt x  - 1 \in U\left( 3 \right)\\
 \to \left[ \begin{array}{l}
\sqrt x  - 1 = 3\\
\sqrt x  - 1 =  - 3\left( l \right)\\
\sqrt x  - 1 = 1\\
\sqrt x  - 1 =  - 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
\sqrt x  = 4\\
\sqrt x  = 2\\
\sqrt x  = 0
\end{array} \right. \to \left[ \begin{array}{l}
x = 16\\
x = 4\\
x = 0
\end{array} \right.
\end{array}\)