Đáp án:
$\begin{array}{l}
Dkxd:x \ge 0;x \ne 1;x \ne \dfrac{1}{4}\\
P = \left( {\dfrac{{2x\sqrt x + x - \sqrt x }}{{x\sqrt x - 1}} - \dfrac{{x + \sqrt x }}{{x - 1}}} \right).\dfrac{{x - 1}}{{2x + \sqrt x - 1}}\\
+ \dfrac{{\sqrt x }}{{2\sqrt x - 1}}\\
= \left( {\dfrac{{2x\sqrt x + x - \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}} - \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}} \right)\\
.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\left( {2\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} + \dfrac{{\sqrt x }}{{2\sqrt x - 1}}\\
= \dfrac{{2x\sqrt x + x - \sqrt x - \sqrt x \left( {x + \sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}.\dfrac{{\sqrt x - 1}}{{2\sqrt x - 1}}\\
+ \dfrac{{\sqrt x }}{{2\sqrt x - 1}}\\
= \dfrac{{x\sqrt x - 2\sqrt x }}{{x + \sqrt x + 1}}.\dfrac{1}{{2\sqrt x - 1}} + \dfrac{{\sqrt x }}{{2\sqrt x - 1}}\\
= \dfrac{{x\sqrt x - 2\sqrt x + \sqrt x \left( {x + \sqrt x + 1} \right)}}{{\left( {2\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{2x\sqrt x + x - \sqrt x }}{{\left( {2\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x \left( {2\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\left( {2\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{x + \sqrt x + 1}}\\
= \dfrac{{x + \sqrt x }}{{x + \sqrt x + 1}}
\end{array}$
