Question

Помогите пожалуйста
5.23;5.24

Answer 1

$(a-b)(a+b)=a^2-b^2 \ sqrt{a}* sqrt{b} = sqrt{ab}$

$5.23)\ sqrt{2+ sqrt{3} } * sqrt{2+ sqrt{2+ sqrt{3} }} * sqrt{2+ sqrt{2+ sqrt{2+ sqrt{3} }} } * \ \ \ * sqrt{2- sqrt{2+ sqrt{2+ sqrt{3} }} }=\ \ \ =sqrt{2- sqrt{2+ sqrt{2+ sqrt{3} }} }*sqrt{2+ sqrt{2+ sqrt{2+ sqrt{3} }} } *sqrt{2+ sqrt{2+ sqrt{3} }}* \ \ * sqrt{2+ sqrt{3} } \ \ 1) sqrt{2- sqrt{2+ sqrt{2+ sqrt{3} }} }*sqrt{2+ sqrt{2+ sqrt{2+ sqrt{3} }} } = \ \ sqrt{2^2-( sqrt{2+ sqrt{2+ sqrt{3} }})^2} = sqrt{4-2- sqrt{2+ sqrt{3} }} =$

$= sqrt{2- sqrt{2+ sqrt{3} } } \ \ 2) sqrt{2- sqrt{2+ sqrt{3} } }*sqrt{2+ sqrt{2+ sqrt{3} }}= sqrt{2^2-( sqrt{2+ sqrt{3}})^2} = \ \ = sqrt{4-2- sqrt{3} } = sqrt{2- sqrt{3} } \ \ 3) sqrt{2- sqrt{3} }* sqrt{2+ sqrt{3} }= sqrt{2^2- (sqrt{3})2} = sqrt{4-3} = sqrt{1} =1 \ \ OTBET: A$

$5.24) \ ( sqrt{2} -1) sqrt{4+ sqrt{9-4 sqrt{2} } } \ \ 1) sqrt{9-4 sqrt{2} }= sqrt{9-2*2 sqrt{2} } = sqrt{9-2 sqrt{8} } = sqrt{8-2 sqrt{8}+1 } = \ \ sqrt{( sqrt{8} -1)^2} = sqrt{8} -1 \ \2)sqrt{4+ sqrt{8} -1 } } = sqrt{3+2 sqrt{2} } = sqrt{2+2 sqrt{2} +1} = sqrt{( sqrt{2} +1)^2} = sqrt{2} +1 \ \ 3) (sqrt{2} -1 )( sqrt{2} +1 )= sqrt{2} ^2-1^2=2-1=1 \ \ OTBET: A$