Đáp án:
\(\dfrac{{\sqrt x + 1}}{{\sqrt x - 1}}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x > 0;x \ne 1\\
A = \left( {\dfrac{{x\sqrt x - 1}}{{x - \sqrt x }} - \dfrac{{x\sqrt x + 1}}{{x + \sqrt x }}} \right):\dfrac{{2\left( {x - 2\sqrt x + 1} \right)}}{{x - 1}}\\
= \left[ {\dfrac{{\left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)}} - \dfrac{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x + 1} \right)}}} \right].\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{2{{\left( {\sqrt x - 1} \right)}^2}}}\\
= \left[ {\dfrac{{x + \sqrt x + 1 - \left( {x - \sqrt x + 1} \right)}}{{\sqrt x }}} \right].\dfrac{{\sqrt x + 1}}{{2\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{x + \sqrt x + 1 - x + \sqrt x - 1}}{{\sqrt x }}.\dfrac{{\sqrt x + 1}}{{2\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{2\sqrt x }}{{\sqrt x }}.\dfrac{{\sqrt x + 1}}{{2\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 1}}
\end{array}\)
