Đáp án:
$\begin{array}{l}
B = \dfrac{2}{{\sqrt 5 - \sqrt 3 }} + \sqrt {6 - 2\sqrt 5 } + \dfrac{{\sqrt {15} - \sqrt {12} }}{{\sqrt 5 - 2}}\\
= \dfrac{{2\left( {\sqrt 5 + \sqrt 3 } \right)}}{{\left( {\sqrt 5 - \sqrt 3 } \right)\left( {\sqrt 5 + \sqrt 3 } \right)}} + \sqrt {5 - 2\sqrt 5 + 1} \\
+ \dfrac{{\sqrt 3 \left( {\sqrt 5 - \sqrt 4 } \right)}}{{\sqrt 5 - \sqrt 4 }}\\
= \dfrac{{2\left( {\sqrt 5 + \sqrt 3 } \right)}}{{5 - 3}} + \sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} + \sqrt 3 \\
= \dfrac{{2\left( {\sqrt 5 + \sqrt 3 } \right)}}{2} + \sqrt 5 - 1 + \sqrt 3 \\
= \sqrt 5 + \sqrt 3 + \sqrt 5 - 1 + \sqrt 3 \\
= 2\sqrt 5 + 2\sqrt 3 - 1
\end{array}$
