#### Answer

$\dfrac{2}{3\sqrt{2}+2\sqrt{3}+7\sqrt{6}+14}$

#### Work Step by Step

Multiplying by the conjugate of the numerator, the rationalized-numerator form of the given expression, $
\dfrac{\sqrt{6}-2}{\sqrt{3}+7}
,$ is
\begin{array}{l}\require{cancel}
\dfrac{\sqrt{6}-2}{\sqrt{3}+7}\cdot\dfrac{\sqrt{6}+2}{\sqrt{6}+2}
\\\\=
\dfrac{(\sqrt{6})^2-(2)^2}{\sqrt{3}(\sqrt{6})+\sqrt{3}(2)+7(\sqrt{6})+7(2)}
\\\\=
\dfrac{6-4}{\sqrt{3(6)}+2\sqrt{3}+7\sqrt{6}+14}
\\\\=
\dfrac{2}{\sqrt{18}+2\sqrt{3}+7\sqrt{6}+14}
\\\\=
\dfrac{2}{\sqrt{9\cdot2}+2\sqrt{3}+7\sqrt{6}+14}
\\\\=
\dfrac{2}{\sqrt{(3)^2\cdot2}+2\sqrt{3}+7\sqrt{6}+14}
\\\\=
\dfrac{2}{3\sqrt{2}+2\sqrt{3}+7\sqrt{6}+14}
.\end{array}