Đáp án:
$\begin{array}{l}
Dkxd:a \ge 0;a \ne 9\\
a)B = \dfrac{{\sqrt a + 1}}{{\sqrt a + 2}} + \dfrac{{\sqrt a + 1}}{{3 - \sqrt a }} - \dfrac{5}{{a - \sqrt a - 6}}\\
= \dfrac{{\left( {\sqrt a + 1} \right)\left( {\sqrt a - 3} \right) - \left( {\sqrt a + 1} \right)\left( {\sqrt a + 2} \right) - 5}}{{\left( {\sqrt a + 2} \right)\left( {\sqrt a - 3} \right)}}\\
= \dfrac{{a - 2\sqrt a - 3 - \left( {a + 3\sqrt a + 2} \right) - 5}}{{\left( {\sqrt a + 2} \right)\left( {\sqrt a - 3} \right)}}\\
= \dfrac{{a - 2\sqrt a - 3 - a - 3\sqrt a - 2 - 5}}{{\left( {\sqrt a + 2} \right)\left( {\sqrt a - 3} \right)}}\\
= \dfrac{{ - 5\sqrt a - 10}}{{\left( {\sqrt a + 2} \right)\left( {\sqrt a - 3} \right)}}\\
= \dfrac{{ - 5\left( {\sqrt a + 2} \right)}}{{\left( {\sqrt a + 2} \right)\left( {\sqrt a - 3} \right)}}\\
= \dfrac{5}{{3 - \sqrt a }}\\
b)B = \dfrac{5}{{3 - \sqrt a }} \in Z\\
\Leftrightarrow \left( {3 - \sqrt a } \right) \in \left\{ { - 5; - 1;1;5} \right\}\\
\Leftrightarrow \sqrt a \in \left\{ {8;4;2; - 2} \right\}\\
Do:\sqrt a \ge 0\\
\Leftrightarrow \sqrt a \in \left\{ {2;4;8} \right\}\\
\Leftrightarrow a \in \left\{ {4;16;64} \right\}\\
Vay\,a \in \left\{ {4;16;64} \right\}
\end{array}$
