#### Answer

$\dfrac{12-3\sqrt{2}}{14}$

#### Work Step by Step

Multiplying by the conjugate of the denominator and using $(a+b)(a-b)=a^2-b^2,$ the given expression, $
\dfrac{3}{\sqrt{2}+4}
,$ simplifies to
\begin{array}{l}\require{cancel}
\dfrac{3}{\sqrt{2}+4}\cdot\dfrac{\sqrt{2}-4}{\sqrt{2}-4}
\\\\=
\dfrac{3\sqrt{2}-12}{(\sqrt{2})^2-(4)^2}
\\\\=
\dfrac{3\sqrt{2}-12}{2-16}
\\\\=
\dfrac{3\sqrt{2}-12}{-14}
\\\\=
\dfrac{12-3\sqrt{2}}{14}
.\end{array}