Đáp án:
$\begin{array}{l}
Dkxd:a > 0\\
\left( {\dfrac{1}{{\sqrt a + 1}} - \dfrac{1}{{a + \sqrt a }}} \right):\dfrac{{\sqrt a - 1}}{{a + 2\sqrt a + 1}}\\
= \left( {\dfrac{1}{{\sqrt a + 1}} - \dfrac{1}{{\sqrt a \left( {\sqrt a + 1} \right)}}} \right):\dfrac{{\sqrt a - 1}}{{{{\left( {\sqrt a + 1} \right)}^2}}}\\
= \dfrac{{\sqrt a - 1}}{{\sqrt a .\left( {\sqrt a + 1} \right)}}.\dfrac{{{{\left( {\sqrt a + 1} \right)}^2}}}{{\sqrt a - 1}}\\
= \dfrac{1}{{\sqrt a }}.\dfrac{{\sqrt a + 1}}{1}\\
= \dfrac{{\sqrt a + 1}}{{\sqrt a }}\\
\text{Vậy}\,Khi:a > 0\\
\left( {\dfrac{1}{{\sqrt a + 1}} - \dfrac{1}{{a + \sqrt a }}} \right):\dfrac{{\sqrt a - 1}}{{a + 2\sqrt a + 1}} = \dfrac{{\sqrt a + 1}}{{\sqrt a }}
\end{array}$
