Answer
$\dfrac{1+5\sqrt{2}}{14}$
Work Step by Step
Multiplying by the conjugate of the denominator, the given expression, $
\dfrac{3+\sqrt{2}}{4\sqrt{2}+2}
,$ is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{3+\sqrt{2}}{4\sqrt{2}+2}\cdot\dfrac{4\sqrt{2}-2}{4\sqrt{2}-2}
\\\\=
\dfrac{(3+\sqrt{2})(4\sqrt{2}-2)}{(4\sqrt{2}+2)(4\sqrt{2}-2)}
\\\\
\text{ (Use $(a+b)(a-b)=a^2-b^2$)}
\\=\dfrac{(3+\sqrt{2})(4\sqrt{2}-2)}{(4\sqrt{2})^2-(2)^2}
\\\\
\text{ (Use FOIL Method)}
\\=
\dfrac{3(4\sqrt{2})+3(-2)+\sqrt{2}(4\sqrt{2})+\sqrt{2}(-2)}{(4\sqrt{2})^2-(2)^2}
\\\\=
\dfrac{12\sqrt{2}-6+4(2)-2\sqrt{2}}{16(2)-4}
\\\\=
\dfrac{12\sqrt{2}-6+8-2\sqrt{2}}{32-4}
\\\\
\text{ (Combine like terms)}
\\=
\dfrac{(-6+8)+\left(12\sqrt{2}-2\sqrt{2}\right)}{28}
\\\\=
\dfrac{2+10\sqrt{2}}{28}
\\\\=
\dfrac{2\left(1+5\sqrt{2}\right)}{28}
\\\\
\text{ (Divide by $2$)}
\\=
\dfrac{\cancel2^1\left(1+5\sqrt{2}\right)}{\cancel{28}^{14}}
\\\\=
\dfrac{1+5\sqrt{2}}{14}
.\end{array}
