Đáp án:
\(\dfrac{{2\left( {\sqrt x + 1} \right)}}{{3\sqrt x }}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x > 0;x \ne \left\{ {1;4} \right\}\\
A = \left( {\dfrac{1}{{\sqrt x - 2}} - \dfrac{1}{{\sqrt x }}} \right):\left( {\dfrac{{\sqrt x - 1}}{{\sqrt x - 2}} - \dfrac{{\sqrt x + 2}}{{\sqrt x + 1}}} \right)\\
= \left[ {\dfrac{{\sqrt x - \sqrt x + 2}}{{\sqrt x \left( {\sqrt x - 2} \right)}}} \right]:\left[ {\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right) - \left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}}} \right]\\
= \dfrac{2}{{\sqrt x \left( {\sqrt x - 2} \right)}}.\dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)}}{{x - 1 - x + 4}}\\
= \dfrac{{2\left( {\sqrt x + 1} \right)}}{{3\sqrt x }}
\end{array}\)
