3√2
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4√3-√8
Hi
$\displaystyle \frac{3\sqrt{2}}{4\sqrt{3}-\sqrt{8}}=\frac{(3\sqrt{2})(4\sqrt{3}+\sqrt{8})}{( 4\sqrt{3}-\sqrt{8})(4\sqrt{3}+\sqrt{8})} = \frac{12\sqrt{6}+12}{(4\sqrt{3})^2-(\sqrt{8})^2}$
$\displaystyle \frac{3\sqrt{2}}{4\sqrt{3}-\sqrt{8}} = \frac{12(\sqrt{6}+1)}{40} = \frac{3(\sqrt{6}+1)}{10}$