Answer
$\dfrac{6-3\sqrt{2}-2\sqrt{5}+\sqrt{10}}{4}$
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{5}}{4+\sqrt{8}}
,$ simplifies to
\begin{array}{l}\require{cancel}
\dfrac{3-\sqrt{5}}{4+\sqrt{8}}\cdot\dfrac{4-\sqrt{8}}{4-\sqrt{8}}
\\\\=
\dfrac{3(4)+3(-\sqrt{8})-\sqrt{5}(4)-\sqrt{5}(-\sqrt{8})}{(4)^2-(\sqrt{8})^2}
\\\\=
\dfrac{12-3\sqrt{8}-4\sqrt{5}+\sqrt{40}}{8}
\\\\=
\dfrac{12-3\sqrt{4\cdot2}-4\sqrt{5}+\sqrt{4\cdot10}}{8}
\\\\=
\dfrac{12-3\sqrt{(2)^2\cdot2}-4\sqrt{5}+\sqrt{(2)^2\cdot10}}{8}
\\\\=
\dfrac{12-3(2)\sqrt{2}-4\sqrt{5}+2\sqrt{10}}{8}
\\\\=
\dfrac{12-6\sqrt{2}-4\sqrt{5}+2\sqrt{10}}{8}
\\\\=
\dfrac{\cancel{2}(6-3\sqrt{2}-2\sqrt{5}+\sqrt{10})}{\cancel{2}(4)}
\\\\=
\dfrac{6-3\sqrt{2}-2\sqrt{5}+\sqrt{10}}{4}
.\end{array}
