Сократите дробь: a) $$\frac{8 + \sqrt{6}}{\sqrt{12} + \sqrt{2}}$$; б) $$\frac{\sqrt{b} + 7}{49 - b}$$.

a) $$\frac{8 + \sqrt{6}}{\sqrt{12} + \sqrt{2}} = \frac{8 + \sqrt{6}}{\sqrt{4 \cdot 3} + \sqrt{2}} = \frac{8 + \sqrt{6}}{2\sqrt{3} + \sqrt{2}} = \frac{(8 + \sqrt{6})(2\sqrt{3} - \sqrt{2})}{(2\sqrt{3} + \sqrt{2})(2\sqrt{3} - \sqrt{2})} = \frac{16\sqrt{3} - 8\sqrt{2} + 2\sqrt{18} - \sqrt{12}}{4 \cdot 3 - 2} = \frac{16\sqrt{3} - 8\sqrt{2} + 2 \sqrt{9 \cdot 2} - \sqrt{4 \cdot 3}}{12 - 2} = \frac{16\sqrt{3} - 8\sqrt{2} + 2 \cdot 3 \sqrt{2} - 2\sqrt{3}}{10} = \frac{14\sqrt{3} - 2\sqrt{2}}{10} = \frac{7\sqrt{3} - \sqrt{2}}{5}$$.

б) $$\frac{\sqrt{b} + 7}{49 - b} = \frac{\sqrt{b} + 7}{7^2 - (\sqrt{b})^2} = \frac{\sqrt{b} + 7}{(7 - \sqrt{b})(7 + \sqrt{b})} = \frac{1}{7 - \sqrt{b}}$$.

Ответ:

a) $$\frac{7\sqrt{3} - \sqrt{2}}{5}$$, б) $$\frac{1}{7 - \sqrt{b}}$$.