2 Сократите дробь: a) 6+√6/√12 + √2; б) √6+7/49-b.

a) $$ \frac{6+\sqrt{6}}{\sqrt{12}+\sqrt{2}} = \frac{6+\sqrt{6}}{\sqrt{4\cdot 3}+\sqrt{2}} = \frac{6+\sqrt{6}}{2\sqrt{3}+\sqrt{2}} = \frac{\sqrt{6}(\sqrt{6}+1)}{\sqrt{2}(2\sqrt{\frac{3}{2}}+1)} = \frac{\sqrt{3}\sqrt{2}(\sqrt{6}+1)}{\sqrt{2}(\sqrt{2}\sqrt{3}+1)} = \frac{\sqrt{3}(\sqrt{6}+1)}{\sqrt{6}+1} = \sqrt{3}$$.

б) $$ \frac{\sqrt{6}+7}{49-b} = \frac{\sqrt{6}+7}{49-b}$$.

Ответ: a) $$\sqrt{3}$$, б) $$\frac{\sqrt{6}+7}{49-b}$$.