Đk: $\left\{ \begin{array}{l} a\ge0 \\ \sqrt a\ne0 \end{array} \right .\Leftrightarrow a>0$
\(\begin{array}{l}
Q = \left( {\dfrac{1}{{\sqrt a + 1}} - \dfrac{1}{{a + \sqrt a }}} \right).\dfrac{{\sqrt a - 1}}{{a + 2\sqrt a + 1}}\\
= \left( {\dfrac{1}{{\sqrt a + 1}} - \dfrac{1}{{\sqrt a (\sqrt a + 1)}}} \right).\dfrac{{\sqrt a - 1}}{{{{(\sqrt a + 1)}^2}}}\\
= \dfrac{{\sqrt a - 1}}{{\sqrt a (\sqrt a + 1)}}.\dfrac{{\sqrt a - 1}}{{{{(\sqrt a + 1)}^2}}}
\end{array}\)
