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

a) Сократим дробь: $$\frac{6 + \sqrt{6}}{\sqrt{12} + \sqrt{2}}$$. $$ \frac{6 + \sqrt{6}}{\sqrt{12} + \sqrt{2}} = \frac{6 + \sqrt{6}}{\sqrt{4 \cdot 3} + \sqrt{2}} = \frac{6 + \sqrt{6}}{2\sqrt{3} + \sqrt{2}} = \frac{\sqrt{6}(\sqrt{6} + 1)}{\sqrt{2}(2\sqrt{\frac{3}{2}} + 1)} $$ $$ \frac{6 + \sqrt{6}}{\sqrt{12} + \sqrt{2}} = \frac{\sqrt{6}(\sqrt{6} + 1)}{\sqrt{2}(2\sqrt{3} + 1)} = \frac{\sqrt{6}(\sqrt{6} + 1)}{\sqrt{2}(\sqrt{4*3} + \sqrt{1})} = \frac{6 + \sqrt{6}}{\sqrt{2}(\sqrt{2}\sqrt{6} + \sqrt{1})} = \frac{6 + \sqrt{6}}{\sqrt{2}(\sqrt{12} + \sqrt{2})} = \frac{\sqrt{6}(\sqrt{6} + 1)}{\sqrt{2}(\sqrt{2}\sqrt{6} + \sqrt{2})} $$ $$ \frac{6 + \sqrt{6}}{2\sqrt{3} + \sqrt{2}} = \frac{(6 + \sqrt{6})(2\sqrt{3} - \sqrt{2})}{(2\sqrt{3} + \sqrt{2})(2\sqrt{3} - \sqrt{2})} = \frac{12\sqrt{3} - 6\sqrt{2} + 2\sqrt{18} - \sqrt{12}}{12 - 2} = \frac{12\sqrt{3} - 6\sqrt{2} + 2 \cdot 3\sqrt{2} - 2\sqrt{3}}{10} = \frac{10\sqrt{3}}{10} = \sqrt{3} $$ б) Сократим дробь: $$\frac{\sqrt{6} + 7}{49 - b}$$. $$ \frac{\sqrt{6} + 7}{49 - b} = \frac{\sqrt{6} + 7}{7^2 - b} $$ Ответ: a) $$\sqrt{3}$$; б) $$\frac{\sqrt{6} + 7}{49 - b}$$