Решите пожалуйста (6) завтра контрольная заранее спасибо

Решите задачу:

$2\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}} = 2\sqrt{6+2\sqrt{5-\sqrt{1+\sqrt{4\cdot12}+12}}} =\\= 2\sqrt{6+2\sqrt{5-\sqrt{1^2+2\sqrt{12}+(\sqrt{12})^2}}} =\\= 2\sqrt{6+2\sqrt{5-\sqrt{(1+\sqrt{12})^2}}} = 2\sqrt{6+2\sqrt{5-(1+\sqrt{12})}} =\\= 2\sqrt{6+2\sqrt{4-\sqrt{12}}} = 2\sqrt{6+2\sqrt{1-\sqrt{4\cdot3}+3}} =\\= 2\sqrt{6+2\sqrt{1^2-2\sqrt{3}+(\sqrt{3})^2}} = 2\sqrt{6+2\sqrt{(1-\sqrt{3})^2}}} =\\= 2\sqrt{6+2|1-\sqrt{3}|} = 2\sqrt{6+2(\sqrt{3}-1)} = 2\sqrt{4+2\sqrt{3}} =$
$=2\sqrt{1+2\sqrt{3}+3} = 2\sqrt{1^2+2\sqrt{3}+(\sqrt{3})^2} = 2\sqrt{(1+\sqrt{3})^2} =\\= 2(1+\sqrt{3}) = 2+2\sqrt{3}$