Đáp án:
$\begin{array}{l}
1)x = 16\left( {tmdk} \right)\\
\Rightarrow \sqrt x = 4\\
\Rightarrow A = \dfrac{1}{{\sqrt x + 1}} = \dfrac{1}{{4 + 1}} = \dfrac{1}{5}\\
2)B = \dfrac{{\sqrt x + 3}}{{\sqrt x - 2}} + \dfrac{{\sqrt x + 2}}{{3 - \sqrt x }} + \dfrac{{\sqrt x + 2}}{{x - 5\sqrt x + 6}}\\
= \dfrac{{\sqrt x + 3}}{{\sqrt x - 2}} - \dfrac{{\sqrt x + 2}}{{\sqrt x - 3}} + \dfrac{{\sqrt x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) - \left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right) + \sqrt x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x - 9 - \left( {x - 4} \right) + \sqrt x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\sqrt x - 3}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{1}{{\sqrt x - 2}}\\
\Rightarrow Q = A:B\\
= \dfrac{1}{{\sqrt x + 1}}:\dfrac{1}{{\sqrt x - 2}}\\
= \dfrac{{\sqrt x - 2}}{{\sqrt x + 1}}\\
3)\left| Q \right| > Q\\
\Rightarrow Q < 0\\
\Rightarrow \dfrac{{\sqrt x - 2}}{{\sqrt x + 1}} < 0\\
\Rightarrow \sqrt x - 2 < 0\\
\Rightarrow \sqrt x < 2\\
\Rightarrow x < 4\\
Vậy\,0 \le x < 4
\end{array}$
