Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
9,\\
a,\\
P = \left( {\dfrac{{x + 2}}{{x\sqrt x  - 1}} - \dfrac{{\sqrt x }}{{x + \sqrt x  + 1}} + \dfrac{1}{{1 - \sqrt x }}} \right):\dfrac{{\sqrt x  - 1}}{2}\\
 = \left( {\dfrac{{x + 2}}{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}} - \dfrac{{\sqrt x }}{{x + \sqrt x  + 1}} - \dfrac{1}{{\sqrt x  - 1}}} \right):\dfrac{{\sqrt x  - 1}}{2}\\
 = \dfrac{{\left( {x + 2} \right) - \sqrt x .\left( {\sqrt x  - 1} \right) - \left( {x + \sqrt x  + 1} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}}:\dfrac{{\sqrt x  - 1}}{2}\\
 = \dfrac{{\left( {x + 2} \right) - \left( {x - \sqrt x } \right) - \left( {x + \sqrt x  + 1} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}}:\dfrac{{\sqrt x  - 1}}{2}\\
 = \dfrac{{x + 2 - x + \sqrt x  - x - \sqrt x  - 1}}{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}}:\dfrac{{\sqrt x  - 1}}{2}\\
 = \dfrac{{ - x + 1}}{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}}.\dfrac{2}{{\sqrt x  - 1}}\\
 = \dfrac{{ - 2.\left( {x - 1} \right)}}{{{{\left( {\sqrt x  - 1} \right)}^2}\left( {x + \sqrt x  + 1} \right)}}\\
 = \dfrac{{ - 2.\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}{{{{\left( {\sqrt x  - 1} \right)}^2}.\left( {x + \sqrt x  + 1} \right)}}\\
 = \dfrac{{ - 2\left( {\sqrt x  + 1} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}}\\
b,\\
P > 2 \Leftrightarrow \dfrac{{ - 2\left( {\sqrt x  + 1} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}} > 2\\
 \Leftrightarrow \dfrac{{\sqrt x  + 1}}{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}} <  - 1\\
 \Leftrightarrow \dfrac{{\sqrt x  + 1}}{{x\sqrt x  - 1}} + 1 < 0\\
 \Leftrightarrow \dfrac{{\sqrt x  + 1 + x\sqrt x  - 1}}{{x\sqrt x  - 1}} < 0\\
 \Leftrightarrow \dfrac{{\sqrt x  + x\sqrt x }}{{x\sqrt x  - 1}} < 0\\
x \ge 0 \Rightarrow \sqrt x  + x\sqrt x  \ge 0 \Rightarrow \left\{ \begin{array}{l}
x \ne 0\\
x\sqrt x  - 1 < 0
\end{array} \right. \Leftrightarrow 0 < x < 1\\
10,\\
C = \dfrac{1}{{2\sqrt x  - 2}} - \dfrac{1}{{2\sqrt x  + 2}} + \dfrac{{\sqrt x }}{{1 - x}}\\
 = \dfrac{1}{{2\left( {\sqrt x  - 1} \right)}} - \dfrac{1}{{2\left( {\sqrt x  + 1} \right)}} - \dfrac{{\sqrt x }}{{x - 1}}\\
 = \dfrac{1}{{2\left( {\sqrt x  - 1} \right)}} - \dfrac{1}{{2\left( {\sqrt x  + 1} \right)}} - \dfrac{{\sqrt x }}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{\left( {\sqrt x  + 1} \right) - \left( {\sqrt x  - 1} \right) - 2\sqrt x }}{{2\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{ - 2\sqrt x  + 2}}{{2\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{ - 2.\left( {\sqrt x  - 1} \right)}}{{2\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{ - 1}}{{\sqrt x  + 1}}\\
b,\\
x = \dfrac{4}{9} \Rightarrow \sqrt x  = \dfrac{2}{3} \Rightarrow C = \dfrac{{ - 1}}{{\dfrac{2}{3} + 1}} =  - \dfrac{3}{5}\\
c,\\
\sqrt x  + 1 \ge 1 > 0,\,\,\forall x \Rightarrow \dfrac{{ - 1}}{{\sqrt x  + 1}} < 0 \Rightarrow C < 0\\
\left| C \right| = \dfrac{1}{3} \Rightarrow C =  - \dfrac{1}{3} \Leftrightarrow \dfrac{{ - 1}}{{\sqrt x  + 1}} =  - \dfrac{1}{3} \Leftrightarrow \sqrt x  + 1 = 3 \Leftrightarrow x = 4
\end{array}\)