те дробь a) 6+√6/√12+√2; б) √6+7/49-6

a) $$\frac{6+\sqrt{6}}{\sqrt{12}+\sqrt{2}}$$

Упростим дробь:

$$\frac{6+\sqrt{6}}{\sqrt{12}+\sqrt{2}} = \frac{6+\sqrt{6}}{\sqrt{4\cdot3}+ \sqrt{2}} = \frac{6+\sqrt{6}}{2\sqrt{3}+\sqrt{2}} = \frac{\sqrt{6}(\sqrt{6}+1)}{\sqrt{2}(2\sqrt{\frac{3}{2}}+1)} = \frac{\sqrt{6}(\sqrt{6}+1)}{\sqrt{2}(\sqrt{6}+1)} = \frac{\sqrt{6}}{\sqrt{2}} = \sqrt{3}$$

б) $$\frac{\sqrt{6}+7}{49-6}$$

Упростим дробь:

$$\frac{\sqrt{6}+7}{49-6} = \frac{\sqrt{6}+7}{43}$$

Ответ: a) $$\sqrt{3}$$; б) $$\frac{\sqrt{6}+7}{43}$$