Đáp án:
$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
a \ge 0\\
\sqrt a \ne 2
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a \ge 0\\
a \ne 4
\end{array} \right.\\
Q = \left( {3 + \dfrac{{a - 2\sqrt a }}{{\sqrt a - 2}}} \right).\left( {3 - \dfrac{{3a + \sqrt a }}{{3\sqrt a + 1}}} \right)\\
= \left( {3 + \dfrac{{\sqrt a \left( {\sqrt a - 2} \right)}}{{\sqrt a - 2}}} \right)\left( {3 - \dfrac{{\sqrt a \left( {3\sqrt a + 1} \right)}}{{3\sqrt a + 1}}} \right)\\
= \left( {3 + \sqrt a } \right)\left( {3 - \sqrt a } \right)\\
= {3^2} - a\\
= 9 - a\\
b)a = \sqrt {6 - 4\sqrt 2 } \left( {tmdk} \right)\\
= \sqrt {4 - 2.2.\sqrt 2 + 2} \\
= \sqrt {{{\left( {2 - \sqrt 2 } \right)}^2}} \\
= 2 - \sqrt 2 \\
\Leftrightarrow Q = 9 - a = 9 - \left( {2 - \sqrt 2 } \right) = 7 + \sqrt 2 \\
c)\left| Q \right| = 4\\
\Leftrightarrow \left[ \begin{array}{l}
Q = 4 = 9 - a \Leftrightarrow a = 5\left( {tm} \right)\\
Q = - 4 = 9 - a \Leftrightarrow a = 13\left( {tm} \right)
\end{array} \right.\\
Vậy\,a = 5/a = 13\\
d)Q = 2a\\
\Leftrightarrow 9 - a = 2a\\
\Leftrightarrow 3a = 9\\
\Leftrightarrow a = 3\left( {tmdk} \right)\\
Vậy\,a = 3
\end{array}$
