Đáp án:

\(\begin{array}{l}
a)\dfrac{{\sqrt x  - 1}}{{\sqrt x  + 1}}\\
b)0 \le x < 9\\
c)x = 4\\
d)x = 0
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)DK:x \ge 0;x \ne 1\\
P = \left[ {\dfrac{1}{{\sqrt x  - 1}} - \dfrac{{2\sqrt x }}{{x\left( {\sqrt x  - 1} \right) + \left( {\sqrt x  - 1} \right)}}} \right]:\left[ {\dfrac{{x + \sqrt x }}{{x\left( {\sqrt x  + 1} \right) + \left( {\sqrt x  - 1} \right)}} + \dfrac{1}{{x + 1}}} \right]\\
 = \dfrac{{x + 1 - 2\sqrt x }}{{\left( {x + 1} \right)\left( {\sqrt x  - 1} \right)}}:\dfrac{{x + \sqrt x  + \sqrt x  + 1}}{{\left( {x + 1} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{{{\left( {\sqrt x  - 1} \right)}^2}}}{{\left( {x + 1} \right)\left( {\sqrt x  - 1} \right)}}.\dfrac{{\left( {x + 1} \right)\left( {\sqrt x  + 1} \right)}}{{{{\left( {\sqrt x  + 1} \right)}^2}}}\\
 = \dfrac{{\sqrt x  - 1}}{{\sqrt x  + 1}}\\
b)P < \dfrac{1}{2}\\
 \to \dfrac{{\sqrt x  - 1}}{{\sqrt x  + 1}} < \dfrac{1}{2}\\
 \to \dfrac{{2\sqrt x  - 2 - \sqrt x  - 1}}{{2\left( {\sqrt x  + 1} \right)}} < 0\\
 \to \dfrac{{\sqrt x  - 3}}{{2\left( {\sqrt x  + 1} \right)}} < 0\\
 \to \sqrt x  - 3 < 0\left( {do:\sqrt x  + 1 > 0\forall x \ge 0} \right)\\
 \to 0 \le x < 9\\
c)P = \dfrac{1}{3}\\
 \to \dfrac{{\sqrt x  - 1}}{{\sqrt x  + 1}} = \dfrac{1}{3}\\
 \to 3\sqrt x  - 3 = \sqrt x  + 1\\
 \to 2\sqrt x  = 4\\
 \to \sqrt x  = 2\\
 \to x = 4\\
d)P = \dfrac{{\sqrt x  - 1}}{{\sqrt x  + 1}} = \dfrac{{\sqrt x  + 1 - 2}}{{\sqrt x  + 1}}\\
 = 1 - \dfrac{2}{{\sqrt x  + 1}}\\
P \in Z \to \dfrac{2}{{\sqrt x  + 1}} \in Z\\
 \to \sqrt x  + 1 \in U\left( 2 \right)\\
 \to \left[ \begin{array}{l}
\sqrt x  + 1 = 2\\
\sqrt x  + 1 = 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 1\left( l \right)\\
x = 0
\end{array} \right.
\end{array}\)