Điều kiện xác định $x\ge 0, x\ne 1$
$\begin{array}{l} P = \left( {\dfrac{1}{{x - \sqrt x }} + \dfrac{1}{{\sqrt x - 1}}} \right):\dfrac{{\sqrt x }}{{x - 2\sqrt x + 1}}\\ P = \left( {\dfrac{1}{{\sqrt x \left( {\sqrt x - 1} \right)}} + \dfrac{1}{{\sqrt x - 1}}} \right).\dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x }}\\ P = \dfrac{{\sqrt x+ 1}}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\sqrt x }} = \dfrac{{ x - 1}}{{ x }} \end{array}$
