Đáp án:
$\begin{array}{l}
a)A = \dfrac{1}{{\sqrt 3 - 1}} - \dfrac{1}{{\sqrt 3 + 1}}\\
= \dfrac{{\sqrt 3 + 1 - \sqrt 3 + 1}}{{\left( {\sqrt 3 - 1} \right)\left( {\sqrt 3 + 1} \right)}}\\
= \dfrac{2}{{3 - 1}}\\
= 1\\
b)A = \dfrac{{2 + \sqrt 3 }}{{2 - \sqrt 3 }} + \dfrac{{2 - \sqrt 3 }}{{2 + \sqrt 3 }}\\
= \dfrac{{{{\left( {2 + \sqrt 3 } \right)}^2} + {{\left( {2 - \sqrt 3 } \right)}^2}}}{{\left( {2 - \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)}}\\
= \dfrac{{4 + 4\sqrt 3 + 3 + 4 - 4\sqrt 3 + 3}}{{4 - 3}}\\
= 14\\
a)A = \dfrac{1}{{\sqrt 3 - \sqrt 2 }} + \dfrac{1}{{\sqrt 3 + \sqrt 2 }}\\
= \dfrac{{\sqrt 3 + \sqrt 2 + \sqrt 3 - \sqrt 2 }}{{3 - 2}}\\
= 2\sqrt 3 \\
B = \dfrac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} + \dfrac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }}\\
= \dfrac{{{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2} + {{\left( {\sqrt 5 + \sqrt 3 } \right)}^2}}}{{5 - 3}}\\
= \dfrac{{8 - 2\sqrt {15} + 8 + 2\sqrt {15} }}{2}\\
= 8
\end{array}$
