Answer
$-\sqrt{6}-3-2\sqrt{2}-2\sqrt{3}$
Work Step by Step
Rationalizing the denominator of $
\dfrac{\sqrt{3}+\sqrt{4}}{\sqrt{2}-\sqrt{3}}
$ results to
\begin{array}{l}
\dfrac{\sqrt{3}+\sqrt{4}}{\sqrt{2}-\sqrt{3}}
\cdot
\dfrac{\sqrt{2}+\sqrt{3}}{\sqrt{2}+\sqrt{3}}
\\\\=
\dfrac{(\sqrt{3})(\sqrt{2})+(\sqrt{3})(\sqrt{3})+(\sqrt{4})(\sqrt{2})+(\sqrt{4})(\sqrt{3})}{(\sqrt{2})^2-(\sqrt{3})^2}
\\\\=
\dfrac{\sqrt{6}+3+\sqrt{8}+\sqrt{12}}{2-3}
\\\\=
\dfrac{\sqrt{6}+3+\sqrt{4\cdot2}+\sqrt{4\cdot3}}{-1}
\\\\=
-\sqrt{6}-3-\sqrt{4\cdot2}-\sqrt{4\cdot3}
\\\\=
-\sqrt{6}-3-2\sqrt{2}-2\sqrt{3}
.\end{array}