Đáp án:
$\begin{array}{l}
A = \dfrac{1}{{3 + \sqrt 3 }} + \dfrac{1}{{2 - \sqrt 2 }}\\
= \dfrac{{3 - \sqrt 3 }}{{{3^2} - 3}} + \dfrac{{2 + \sqrt 2 }}{{{2^2} - 2}}\\
= \dfrac{{3 - \sqrt 3 }}{6} + \dfrac{{2 + \sqrt 2 }}{2}\\
= \dfrac{{3 - \sqrt 3 + 6 + 3\sqrt 2 }}{6}\\
= \dfrac{{9 + 3\sqrt 2 - \sqrt 3 }}{6}\\
B = \dfrac{3}{{3\sqrt 2 - 4}} - \dfrac{2}{{3\sqrt 2 + 4}}\\
= \dfrac{{3\left( {3\sqrt 2 + 4} \right) - 2\left( {3\sqrt 2 - 4} \right)}}{{{{\left( {3\sqrt 2 } \right)}^2} - {4^2}}}\\
= \dfrac{{9\sqrt 2 + 12 - 6\sqrt 2 + 8}}{{18 - 16}}\\
= \dfrac{{3\sqrt 2 + 20}}{2}
\end{array}$
