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