Question

Помогите с 192 номером, прошу умоляю очень надо, прошууу отзовитесь

Answer 1

решение во вложении.............

Answer 2

$frac{2+ sqrt{6}}{2 sqrt{2}+2 sqrt{3}- sqrt{6}-2}= frac{2+ sqrt{6}}{sqrt{2}*2+ sqrt{2}* sqrt{2}* sqrt{3}- sqrt{6}-2}= frac{2+ sqrt{6}}{sqrt{2}*(2+ sqrt{2}*sqrt{3})- (sqrt{6}+2)}=$

$= frac{2+ sqrt{6}}{sqrt{2}*(2+ sqrt{2*3})-1*(sqrt{6}+2)}= frac{2+ sqrt{6}}{sqrt{2}*(2+ sqrt{6})-1*(2+sqrt{6})}= frac{1*(2+ sqrt{6})}{(sqrt{2}-1)*(2+ sqrt{6})}=$

$= frac{1}{sqrt{2}-1}=frac{1*( sqrt{2}+1)}{(sqrt{2}-1)*( sqrt{2}+1)}=frac{1*( sqrt{2}+1)}{(sqrt{2})^2-1^2}=frac{sqrt{2}+1}{2-1}=1+sqrt{2}$

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$sqrt{ frac{2+ sqrt{2} }{2- sqrt{2} }* sqrt{ frac{2-sqrt{2} }{2+sqrt{2} }} } = sqrt{( sqrt{frac{2+ sqrt{2} }{2- sqrt{2} }} )^2* sqrt{ frac{ 2-sqrt{2} }{2+sqrt{2} }} } =$

$=sqrt{sqrt{frac{2+ sqrt{2} }{2- sqrt{2} }}*sqrt{frac{2+ sqrt{2} }{2- sqrt{2} }}* sqrt{ frac{ 2-sqrt{2} }{2+sqrt{2} }} } =sqrt{sqrt{frac{2+ sqrt{2} }{2- sqrt{2} }*frac{2+ sqrt{2} }{2- sqrt{2} }*frac{ 2-sqrt{2} }{2+sqrt{2} }} } =$

$=sqrt{sqrt{frac{2+ sqrt{2} }{2- sqrt{2}}}} =sqrt{sqrt{frac{(2+ sqrt{2})(2+ sqrt{2})}{(2- sqrt{2})(2+ sqrt{2} )}}}} = sqrt{sqrt{frac{(2+ sqrt{2})^2}{2^2- (sqrt{2})^2}}}} =$

$=sqrt{sqrt{frac{(2+ sqrt{2})^2}{4- 2}}}} = sqrt{ frac{2+ sqrt{2} }{ sqrt{2} } } = sqrt{ frac{ sqrt{2} ( sqrt{2}+1 )}{ sqrt{2} } } = sqrt{ sqrt{2}+1 }$