\(=\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{4}\right)}{\sqrt{5}-2}-\dfrac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=\dfrac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}-\dfrac{2+\sqrt{3}}{4-3}=\sqrt{3}-\left(2+\sqrt{3}\right)=-2\)
$\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\\
