Question

задание на фото:::::

Answer 1

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

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

Answer 2

Надеюсь, достаточно подробно)

Answer 3

Спасибо ^^

Answer 4

Пожалуйста))