Đáp án:
$\begin{array}{l}
1)\\
P = \left( {\frac{{3\sqrt a }}{{\sqrt a + 4}} + \frac{{\sqrt a }}{{\sqrt a - 4}} + \frac{{4\left( {a + 2} \right)}}{{16 - a}}} \right):\left( {1 - \frac{{2\sqrt a + 5}}{{\sqrt a + 4}}} \right)\\
= \frac{{3\sqrt a \left( {\sqrt a - 4} \right) + \sqrt a \left( {\sqrt a + 4} \right) - 4a - 8}}{{\left( {\sqrt a - 4} \right)\left( {\sqrt a + 4} \right)}}:\frac{{\sqrt a + 4 - 2\sqrt a - 5}}{{\sqrt a + 4}}\\
= \frac{{3a - 12\sqrt a + a + 4\sqrt a - 4a - 8}}{{\left( {\sqrt a - 4} \right)\left( {\sqrt a + 4} \right)}}.\frac{{\left( {\sqrt a + 4} \right)}}{{ - \sqrt a - 1}}\\
= \frac{{ - 8\sqrt a - 8}}{{\left( {\sqrt a - 4} \right).\left( { - \sqrt a - 1} \right)}}\\
= \frac{8}{{\sqrt a - 4}}\\
2)a \ge 0;a \ne 16\\
P = - 3\\
\Rightarrow \frac{8}{{\sqrt a - 4}} = - 3\\
\Rightarrow 8 = - 3\sqrt a + 12\\
\Rightarrow 3\sqrt a = 4\\
\Rightarrow \sqrt a = \frac{4}{3}\\
\Rightarrow a = \frac{{16}}{9}\left( {tmdk} \right)\\
3)a \ge 0;a \ne 16\\
P = \frac{8}{{\sqrt a - 4}}
\end{array}$
P là số nguyên tố thì 8 chia hết cho mẫu
=> P = 2
$\begin{array}{l}
\Rightarrow \sqrt a - 4 = 4\\
\Rightarrow \sqrt a = 8\\
\Rightarrow a = 64\left( {tm} \right)
\end{array}$
