#### Answer

$\dfrac{3\sqrt{2}-\sqrt{3}}{5}$

#### Work Step by Step

Multiplying by the conjugate of the denominator and using $(a+b)(a-b)=a^2-b^2$, the simplified form of the given expression, $
\dfrac{\sqrt{6}}{\sqrt{12}+\sqrt{2}}
,$ is
\begin{array}{l}\require{cancel}
\dfrac{\sqrt{6}}{\sqrt{12}+\sqrt{2}}\cdot\dfrac{\sqrt{12}-\sqrt{2}}{\sqrt{12}-\sqrt{2}}
\\\\=
\dfrac{\sqrt{6(12)}-\sqrt{6(2)}}{(\sqrt{12})^2-(\sqrt{2})^2}
\\\\=
\dfrac{\sqrt{6(6\cdot2)}-\sqrt{(2\cdot3)(2)}}{12-2}
\\\\=
\dfrac{\sqrt{(6)^2\cdot2}-\sqrt{(2)^2\cdot3}}{10}
\\\\=
\dfrac{6\sqrt{2}-2\sqrt{3}}{10}
\\\\=
\dfrac{2(3\sqrt{2}-\sqrt{3})}{10}
\\\\=
\dfrac{\cancel{2}(3\sqrt{2}-\sqrt{3})}{\cancel{2}(5)}
\\\\=
\dfrac{3\sqrt{2}-\sqrt{3}}{5}
.\end{array}