sqrt=radical
^=ridicarea la putere
sqrt[(1-sqrt7)^2]+sqrt(16-6sqrt7)
Ne ocupam prima data de al doilea radical
6sqrt7=2ab/:2=>3sqrt7=cb. Fie c=3 si b=sqrt7
Am facut asa pentru a afla cum se descompune relatia de sub radical. Si acum verific daca am dat valorile lui c si b bine:
3^2+sqrt7=16=>9+7=16(A)
Sqrt[(1-sqrt7)^2]=|1-sqrt7|
1 |1-sqrt7|=sqrt7-1
Sqrt(16-6sqrt7)=sqrt[(3-sqrt7)^2]=|3-sqrt7|
3>sqrt7=>|3-sqrt7|=3-sqrt7
a=sqrt7-1+3-sqrt7
a=-1+3=>a=2
