Задание внутри. Выберу лучший ответ.

a)[latex] \frac{5}{3 \sqrt{10} } = \frac{5 \sqrt{10} }{3 \sqrt{10} \sqrt{10} } = \frac{5 \sqrt{10} }{3*10} = \frac{ \sqrt{10} }{6} [/latex] b) [latex] \frac{8}{ \sqrt{6}+ \sqrt{2} } = \frac{8 (\sqrt{6}- \sqrt{2} ) }{( \sqrt{6}+ \sqrt{2})( \sqrt{6}- \sqrt{2})}= \frac{8 (\sqrt{6}- \sqrt{2} )}{6-2}= \frac{8 (\sqrt{6}- \sqrt{2} )}{4} = 2( \sqrt{6} - \sqrt{2} )[/latex]

[latex]a)\ \frac{5}{3\sqrt{10}}=\frac{5\sqrt{10}}{3\sqrt{10}\bullet\sqrt{10}}=\frac{5\sqrt{10}}{3(\sqrt{10})^2}=\frac{5\sqrt{10}}{3\bullet10}=\frac{\sqrt{10}}{6};\\\\b)\ \frac{8}{\sqrt{6}+\sqrt2}=\frac{8(\sqrt6-\sqrt2)}{(\sqrt6+\sqrt2)(\sqrt6-\sqrt2)}=\frac{8(\sqrt6-\sqrt2)}{(\sqrt{6})^2-(\sqrt{2})^2}=\frac{8(\sqrt6-\sqrt2)}{6-2}=\\\\=2(\sqrt{3\bullet2}-\sqrt2)=2(\sqrt2\sqrt3-\sqrt2)=2\sqrt2(\sqrt3-1).[/latex]