Đáp án:
$\begin{array}{l}
A = \dfrac{{\sqrt x + 3}}{{\sqrt x }} + \dfrac{{3\sqrt x + 9}}{{x - 3\sqrt x }}\\
= \dfrac{{\sqrt x + 3}}{{\sqrt x }} + \dfrac{{3\sqrt x + 9}}{{\sqrt x \left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) + 3\sqrt x + 9}}{{\sqrt x \left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x - 9 + 3\sqrt x + 9}}{{\sqrt x \left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x + 3\sqrt x }}{{\sqrt x \left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\sqrt x + 3}}{{\sqrt x - 3}}\\
B = \dfrac{{\sqrt x + 2}}{{\sqrt x }} + \dfrac{{3\sqrt x + 4}}{{x - 2\sqrt x }}\\
= \dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right) + 3\sqrt x + 4}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x - 4 + 3\sqrt x + 4}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x + 3\sqrt x }}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\sqrt x + 3}}{{\sqrt x - 2}}
\end{array}$
