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

a) $$\frac{6+\sqrt{8}}{\sqrt{12}+\sqrt{2}} = \frac{6+\sqrt{4\cdot2}}{\sqrt{4\cdot3}+ \sqrt{2}} = \frac{6+2\sqrt{2}}{2\sqrt{3}+\sqrt{2}} = \frac{2(3+\sqrt{2})}{2\sqrt{3}+\sqrt{2}} = \frac{2(3+\sqrt{2})}{\sqrt{2}(2\frac{\sqrt{3}}{\sqrt{2}}+1)} = \frac{2(3+\sqrt{2})}{\sqrt{2}(\sqrt{6}+1)} = \frac{2(3+\sqrt{2})(\sqrt{6}-1)}{\sqrt{2}(\sqrt{6}+1)(\sqrt{6}-1)} = \frac{2(3\sqrt{6}-3+\sqrt{12}-\sqrt{2})}{\sqrt{2}(6-1)} = \frac{2(3\sqrt{6}-3+2\sqrt{3}-\sqrt{2})}{5\sqrt{2}} = \frac{2(3\sqrt{6}-3+2\sqrt{3}-\sqrt{2})\sqrt{2}}{5\sqrt{2}\sqrt{2}} = \frac{2(3\sqrt{12}-3\sqrt{2}+2\sqrt{6}-2)}{5\cdot2} = \frac{2(3\cdot2\sqrt{3}-3\sqrt{2}+2\sqrt{6}-2)}{10} = \frac{12\sqrt{3}-6\sqrt{2}+4\sqrt{6}-4}{10} = \frac{6\sqrt{3}-3\sqrt{2}+2\sqrt{6}-2}{5}$$. 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) $$\frac{6\sqrt{3}-3\sqrt{2}+2\sqrt{6}-2}{5}$$; б) $$\frac{1}{7-\sqrt{b}}$$.