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