Giải thích các bước giải:
\(\begin{array}{l}
P = \left( {\dfrac{{x + 2}}{{x\sqrt x - 1}} + \dfrac{{\sqrt x }}{{x + \sqrt x + 1}} + \dfrac{1}{{1 - \sqrt x }}} \right):\dfrac{{\sqrt x - 1}}{2}\\
= \dfrac{{x + 2 + \sqrt x \left( {\sqrt x - 1} \right) - \left( {x + \sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{2}{{\sqrt x - 1}}\\
= \dfrac{{x + 2 + x - \sqrt x - x - \sqrt x - 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{2}{{\sqrt x - 1}}\\
= \dfrac{{x - 2\sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{2}{{\sqrt x - 1}}\\
= \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{2}{{\sqrt x - 1}}\\
= \dfrac{2}{{x + \sqrt x + 1}}\\
Do\,x + \sqrt x + 1 = {\left( {\sqrt x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} > 0\\
\Rightarrow P = \dfrac{2}{{x + \sqrt x + 1}} > 0\,\,voi\,x \ge 0;x \ne 1
\end{array}\)
