SE APLICA FORMULA RADICALILOR COMPUSI
√(a+/-√b)=√(a+√(a²-b))+/-√(a-√(a²-b))
Deci √(4-2√3)=√4-√12)=√(4+2)-√4-2)=√6-√2
√(4+2√3)=√4+√12)=√(4+2)+√4-2)=√6+√2
Ecuatia devine (2x+1)(√6-√2)+(2x-1)(√6+√2)=2√3(4-x√3)+10
2x√6+√6-2x√2-√2+2x√6+2x√2-√6-√2=2√3(4-x√3)+10
4x√6-2√2=2√3(4-x√3)+10
x(4√6+2*3)=8√3+2√2+10
x=(8√3+2√2+10)/(4√6+2*3)=(4√3+√2+5)/(2√6+3)
Sper sa nu fi gresit la calcule, dar am dat modul de rezolvare!
