Решение задания 3

1. а) $$\frac{6+\sqrt{6}}{\sqrt{30}+\sqrt{5}} = \frac{6+\sqrt{6}}{\sqrt{5}(\sqrt{6}+1)} = \frac{\sqrt{6}(\sqrt{6}+1)}{\sqrt{5}(\sqrt{6}+1)} = \frac{\sqrt{6}}{\sqrt{5}} = \frac{\sqrt{6}\cdot \sqrt{5}}{\sqrt{5}\cdot \sqrt{5}} = \frac{\sqrt{30}}{5}$$.
2. б) $$\frac{9-a}{3+\sqrt{a}} = \frac{(3-\sqrt{a})(3+\sqrt{a})}{3+\sqrt{a}} = 3-\sqrt{a}$$.

Ответ: а) $$\frac{\sqrt{30}}{5}$$; б) $$3-\sqrt{a}$$.