$\begin{array}{l} \left( {\dfrac{1}{{\sqrt a – 1}} – \dfrac{1}{{\sqrt a }}} \right):\left( {\dfrac{{\sqrt a + 1}}{{\sqrt a – 2}} – \dfrac{{\sqrt a + 2}}{{\sqrt a – 1}}} \right)\\ = \dfrac{{\sqrt a – \left( {\sqrt a – 1} \right)}}{{\sqrt a \left( {\sqrt a – 1} \right)}}:\dfrac{{\left( {\sqrt a + 1} \right)\left( {\sqrt a – 1} \right) – \left( {\sqrt a + 2} \right)\left( {\sqrt a – 2} \right)}}{{\left( {\sqrt a – 1} \right)\left( {\sqrt a – 2} \right)}}\\ = \dfrac{1}{{\sqrt a \left( {\sqrt a – 1} \right)}}.\dfrac{{\left( {\sqrt a – 1} \right)\left( {\sqrt a – 2} \right)}}{{a – 1 – \left( {a – 4} \right)}}\\ = \dfrac{1}{{\sqrt a }}.\dfrac{{\sqrt a – 2}}{3}\\ = \dfrac{{\sqrt a – 2}}{{3\sqrt a }} \end{array}$
Đáp án:
$\begin{array}{l}
\left( {\dfrac{1}{{\sqrt a – 1}} – \dfrac{1}{{\sqrt a }}} \right):\left( {\dfrac{{\sqrt a + 1}}{{\sqrt a – 2}} – \dfrac{{\sqrt a + 2}}{{\sqrt a – 1}}} \right)\\
= \dfrac{{\sqrt a – \left( {\sqrt a – 1} \right)}}{{\sqrt a \left( {\sqrt a – 1} \right)}}:\dfrac{{\left( {\sqrt a + 1} \right)\left( {\sqrt a – 1} \right) – \left( {\sqrt a + 2} \right)\left( {\sqrt a – 2} \right)}}{{\left( {\sqrt a – 1} \right)\left( {\sqrt a – 2} \right)}}\\
= \dfrac{1}{{\sqrt a \left( {\sqrt a – 1} \right)}}.\dfrac{{\left( {\sqrt a – 1} \right)\left( {\sqrt a – 2} \right)}}{{a – 1 – \left( {a – 4} \right)}}\\
= \dfrac{1}{{\sqrt a }}.\dfrac{{\sqrt a – 2}}{3}\\
= \dfrac{{\sqrt a – 2}}{{3\sqrt a }}
\end{array}$