Đáp án:
\(\dfrac{{\left( {x - \sqrt x + 1} \right)\left( {\sqrt x - 2} \right)}}{{x\left( {\sqrt x + 2} \right)}}\)
Giải thích các bước giải:
\(\begin{array}{l}
D = \left( {\dfrac{1}{{\sqrt x - 1}} - \dfrac{1}{x}} \right):\left( {\dfrac{{\sqrt x + 2}}{{\sqrt x - 2}} - \dfrac{{\sqrt x + 2}}{{\sqrt x - 1}}} \right)\\
= \left( {\dfrac{{x - \sqrt x + 1}}{{x\left( {\sqrt x - 1} \right)}}} \right):\left[ {\dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right) - \left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 1} \right)}}} \right]\\
= \dfrac{{x - \sqrt x + 1}}{{x\left( {\sqrt x - 1} \right)}}.\dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 1} \right)}}{{x + \sqrt x - 2 - x + 4}}\\
= \dfrac{{x - \sqrt x + 1}}{{x\left( {\sqrt x - 1} \right)}}.\dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 1} \right)}}{{\sqrt x + 2}}\\
= \dfrac{{\left( {x - \sqrt x + 1} \right)\left( {\sqrt x - 2} \right)}}{{x\left( {\sqrt x + 2} \right)}}
\end{array}\)
