Answer
$\dfrac{3\sqrt{42}-4\sqrt{15}}{23}$
Work Step by Step
Multiplying by the conjugate of the denominator and using $(a+b)(a-b)=a^2-b^2$, the simplifed form of given expression, $
\dfrac{\sqrt{6}}{3\sqrt{7}+2\sqrt{10}}
,$ is
\begin{array}{l}\require{cancel}
\dfrac{\sqrt{6}}{3\sqrt{7}+2\sqrt{10}}\cdot\dfrac{3\sqrt{7}-2\sqrt{10}}{3\sqrt{7}-2\sqrt{10}}
\\\\=
\dfrac{3\sqrt{6(7)}-2\sqrt{6(10)}}{(3\sqrt{7})^2-(2\sqrt{10})^2}
\\\\=
\dfrac{3\sqrt{42}-2\sqrt{60}}{9(7)-4(10)}
\\\\=
\dfrac{3\sqrt{42}-2\sqrt{4\cdot15}}{63-40}
\\\\=
\dfrac{3\sqrt{42}-2\sqrt{(2)^2\cdot15}}{23}
\\\\=
\dfrac{3\sqrt{42}-2(2)\sqrt{15}}{23}
\\\\=
\dfrac{3\sqrt{42}-4\sqrt{15}}{23}
.\end{array}
