Задание на фото:::::

Решите задачу:

$=\cfrac{\sqrt{6}\cdot \sqrt{2-\sqrt{3}}\cdot (\sqrt{15}+\sqrt{5}-\sqrt{3}-1)(\sqrt{5}+1)}{2\sqrt{2} \cdot \sqrt{2+\sqrt{3}} \cdot (1-\sqrt{3})} = \\\\\\ = \cfrac{\sqrt{6(2-\sqrt{3})}\cdot (\sqrt{5} \cdot \sqrt{3}+\sqrt{5}-(\sqrt{3}+1))(\sqrt{5}+1)}{2\sqrt{2(2+\sqrt{3})} \cdot (1-\sqrt{3})} = \\\\\\ = \cfrac{\sqrt{12-6\sqrt{3}}\cdot (\sqrt{5} (\sqrt{3}+1)-(\sqrt{3}+1))(\sqrt{5}+1)}{2\sqrt{4+2\sqrt{3}} \cdot (1-\sqrt{3})} =$

$= \cfrac{\sqrt{9-6\sqrt{3}+3}\cdot (\sqrt{3}+1)(\sqrt{5} -1)(\sqrt{5}+1)}{2\sqrt{1+2\sqrt{3}+3} \cdot (1-\sqrt{3})} = \\\\\\ = \cfrac{\sqrt{(3-\sqrt{3})^2}\cdot (\sqrt{3}+1)(5-1)}{2\sqrt{(1+\sqrt{3})^2} \cdot (1-\sqrt{3})} = \cfrac{(3-\sqrt{3}) (\sqrt{3}+1) \cdot4}{2(1+\sqrt{3})(1-\sqrt{3})} = \\\\\\ =\cfrac{(3\sqrt{3}+3-3-\sqrt{3}) \cdot2}{1-3} = \cfrac{2\sqrt{3} \cdot2}{-2} =-2\sqrt{3}$

Задание ** фото:::::