Đáp án:
$\begin{array}{l}
2)a)Dkxd:x \ge 0;x \ne 9\\
Q = \dfrac{{2\sqrt x }}{{\sqrt x + 3}} + \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}} + \dfrac{{3 - 11\sqrt x }}{{9 - x}}\\
= \dfrac{{2\sqrt x \left( {\sqrt x - 3} \right) + \left( {\sqrt x + 1} \right)\left( {\sqrt x + 3} \right) - 3 + 11\sqrt x }}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{2x - 6\sqrt x + x + 4\sqrt x + 3 - 3 + 11\sqrt x }}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{3x + 9\sqrt x }}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{3\sqrt x \left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{3\sqrt x }}{{\sqrt x - 3}}\\
b)\left| Q \right| = Q\\
\Rightarrow Q \ge 0\\
\Rightarrow \dfrac{{3\sqrt x }}{{\sqrt x - 3}} \ge 0\\
\Rightarrow \sqrt x - 3 \ge 0\\
\Rightarrow \sqrt x \ge 3\\
\Rightarrow x \ge 9\\
Vậy\,x > 9\left( {do:x \ne 9} \right)
\end{array}$
