Question

Объясните, как решать, пожалуйста

Answer 1

$(a-b)^2=a^2-2ab+b^2 \ (a+b)^2=a^2-2ab+b^2$



$sqrt{( sqrt{7-2 sqrt{10} }+ sqrt{2})*2 sqrt{5} } =sqrt{10} \ \ \1)~ 7-2 sqrt{10} =5+2-2 sqrt{10} =5-2 sqrt{5} *sqrt{2} +2= \ = ( sqrt{5} )^2-2 sqrt{5} *sqrt{2} +(sqrt{2})^2 =( sqrt{5} - sqrt{2})^2 \ 2)~sqrt{( sqrt{5} - sqrt{2})^2} =sqrt{5} - sqrt{2}\ 3)~ ( sqrt{5} - sqrt{2}+ sqrt{2})*2 sqrt{5} =2 sqrt{5}* sqrt{5} -2 sqrt{5}* sqrt{2} +2 sqrt{5}*sqrt{2} = \ =2*5 -2 sqrt{5*2} +2 sqrt{5*2}=10-2 sqrt{10} +2 sqrt{10} =10 \ 4)~ sqrt{10} approx 3.162277$



$sqrt{( sqrt{16-6 sqrt{7} }+ sqrt{7})*3 }= sqrt{( sqrt{3^2-3*2 sqrt{7} +7}+ sqrt{7})*3 }= \ =sqrt{( sqrt{(3-sqrt{7})^2 }+ sqrt{7})*3 }= sqrt{(3 -sqrt{7}+ sqrt{7})*3 }= sqrt{(3*3 }= \ = sqrt{9} =3$



$sqrt{( sqrt{8+2 sqrt{15} } -sqrt{8-2 sqrt{15} } )*2+7} = \ = sqrt{( sqrt{5+2 sqrt{5*3} +3} -sqrt{5-2 sqrt{5*3}+3 } )*2+7} \ = sqrt{( sqrt{( sqrt{5} + sqrt{3})^2} -sqrt{( sqrt{5} -sqrt{3})^2} )*2+7} = \ =sqrt{(sqrt{5} + sqrt{3} -sqrt{5} +sqrt{3} )*2+7} = \ = sqrt{2 sqrt{3} *2+7} = sqrt{4 sqrt{3}+7 }= sqrt{2^2+2*2sqrt{3}+3} = \ =sqrt{(2+ sqrt{3})^2 } =2+ sqrt{3}$