Question

Помогите пожалуйста


Вычислите:

Answer 1

$sqrt{2+ sqrt{3} } sqrt{2+ sqrt{2+ sqrt{3} } } sqrt{2- sqrt{2+ sqrt{3} } } = \ = sqrt{2+ sqrt{3} } sqrt{(2+ sqrt{2+ sqrt{3} })(2- sqrt{2+ sqrt{3} } )} = \ = sqrt{2+ sqrt{3} } sqrt{4- 2- sqrt{3}} =sqrt{2+ sqrt{3} } sqrt{ 2- sqrt{3}} = \ =sqrt{(2+ sqrt{3})( 2- sqrt{3})} =sqrt{4- 3} = sqrt{1} =1$

$( sqrt{11+6 sqrt{2} }+ sqrt{11-6 sqrt{2} })^2= \ =11+6 sqrt{2} + 11-6 sqrt{2} +2 sqrt{(11+6 sqrt{2})(11-6 sqrt{2})} = \ =22 +2 sqrt{121-72} =22 +2 sqrt{49} =22 +14=36$

Answer 2

Решение на картинке.