Answer
$\dfrac{5}{7+3\sqrt{6}}$
Work Step by Step
Multiplying by the conjugate of the numerator, then the rationalized-numerator form of the given expression, $
\dfrac{\sqrt{8}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}
,$ is
\begin{array}{l}\require{cancel}
\dfrac{\sqrt{8}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}\cdot\dfrac{\sqrt{8}+\sqrt{3}}{\sqrt{8}+\sqrt{3}}
\\\\=
\dfrac{(\sqrt{8})^2-(\sqrt{3})^2}{\sqrt{2}(\sqrt{8})+\sqrt{2}(\sqrt{3})+\sqrt{3}(\sqrt{8})+\sqrt{3}(\sqrt{3})}
\\\\=
\dfrac{8-3}{\sqrt{2(8)}+\sqrt{2(3)}+\sqrt{3(8)}+\sqrt{3(3)}}
\\\\=
\dfrac{5}{\sqrt{16}+\sqrt{6}+\sqrt{24}+\sqrt{9}}
\\\\=
\dfrac{5}{\sqrt{(4)^2}+\sqrt{6}+\sqrt{4\cdot6}+\sqrt{(3)^2}}
\\\\=
\dfrac{5}{\sqrt{(4)^2}+\sqrt{6}+\sqrt{(2)^2\cdot6}+\sqrt{(3)^2}}
\\\\=
\dfrac{5}{4+\sqrt{6}+2\sqrt{6}+3}
\\\\=
\dfrac{5}{7+3\sqrt{6}}
.\end{array}