Решение задания 4

1. а) $$\frac{1}{2\sqrt{5}} = \frac{1\cdot \sqrt{5}}{2\sqrt{5}\cdot \sqrt{5}} = \frac{\sqrt{5}}{2\cdot 5} = \frac{\sqrt{5}}{10}$$.
2. б) $$\frac{8}{\sqrt{7}-1} = \frac{8(\sqrt{7}+1)}{(\sqrt{7}-1)(\sqrt{7}+1)} = \frac{8(\sqrt{7}+1)}{7-1} = \frac{8(\sqrt{7}+1)}{6} = \frac{4(\sqrt{7}+1)}{3}$$.

Ответ: а) $$\frac{\sqrt{5}}{10}$$; б) $$\frac{4(\sqrt{7}+1)}{3}$$.