Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
P = \dfrac{{x + 2}}{{x\sqrt x  - 1}} + \dfrac{{\sqrt x  + 1}}{{x + \sqrt x  + 1}} - \dfrac{{\sqrt x  + 1}}{{x - 1}}\\
 = \dfrac{{x + 2}}{{{{\sqrt x }^3} - {1^3}}} + \dfrac{{\sqrt x  + 1}}{{x + \sqrt x  + 1}} - \dfrac{{\sqrt x  + 1}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}\\
 = \dfrac{{x + 2}}{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}} + \dfrac{{\sqrt x  + 1}}{{x + \sqrt x  + 1}} - \dfrac{1}{{\sqrt x  - 1}}\\
 = \dfrac{{\left( {x + 2} \right) + \left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right) - 1.\left( {x + \sqrt x  + 1} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}}\\
 = \dfrac{{x + 2 + x - 1 - x - \sqrt x  - 1}}{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}}\\
 = \dfrac{{x - \sqrt x }}{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}}\\
 = \dfrac{{\sqrt x \left( {\sqrt x  - 1} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {x + \sqrt x  + 1} \right)}}\\
 = \dfrac{{\sqrt x }}{{x + \sqrt x  + 1}}\\
b,\\
P - \dfrac{1}{3} = \dfrac{{\sqrt x }}{{x + \sqrt x  + 1}} - \dfrac{1}{3}\\
 = \dfrac{{3\sqrt x  - \left( {x + \sqrt x  + 1} \right)}}{{3.\left( {x + \sqrt x  + 1} \right)}}\\
 = \dfrac{{ - x + 2\sqrt x  - 1}}{{3.\left( {x + \sqrt x  + 1} \right)}}\\
 = \dfrac{{ - \left( {x - 2\sqrt x  + 1} \right)}}{{3.\left( {x + \sqrt x  + 1} \right)}}\\
 = \dfrac{{ - {{\left( {\sqrt x  - 1} \right)}^2}}}{{3.\left( {x + \sqrt x  + 1} \right)}} < 0,\,\,\,\forall x > 0,x \ne 1\\
 \Rightarrow P < \dfrac{1}{3},\,\,\,\,\forall x > 0,x \ne 1
\end{array}\)