Сократите дробь: a) $$\frac{6 + \sqrt{6}}{\sqrt{12} + \sqrt{2}}$$; б) $$\frac{\sqrt{b} + 7}{49 - b}$$.

a) $$\frac{6 + \sqrt{6}}{\sqrt{12} + \sqrt{2}} = \frac{6 + \sqrt{6}}{\sqrt{4 \cdot 3} + \sqrt{2}} = \frac{6 + \sqrt{6}}{2\sqrt{3} + \sqrt{2}} = \frac{\sqrt{6}(\sqrt{6} + 1)}{\sqrt{2}(2\sqrt{\frac{3}{2}} + 1)} = \frac{\sqrt{3}(\sqrt{6} + 1)}{(\sqrt{3} + 1)} = \frac{\sqrt{6}(\sqrt{6} + 1)}{\sqrt{2}(\sqrt{6} + 1)} = \frac{\sqrt{6}}{\sqrt{2}} = \sqrt{3}$$. б) $$\frac{\sqrt{b} + 7}{49 - b} = \frac{\sqrt{b} + 7}{(7 - \sqrt{b})(7 + \sqrt{b})} = \frac{1}{7 - \sqrt{b}}$$. Ответ: a) $$\sqrt{3}$$; б) $$\frac{1}{7 - \sqrt{b}}$$.