Question

Exercitiu 2 de la subiectul 3 va rog

Answer 1

√(2-√5)²+x√(9-4√5)=x√(2+√5)²-√(9+4√5)=
2-√5<0
√(2-√5)²=I2-√5I=√5-2
formula radicali compusi
√(9-4√5)=√(9-√80)=√[9+√(81-80)/2-√[9+√(81-80)/2=√(9+√1)/2-√(9-√1)/2=
=√(9+1)/2-√(9-1)/2=√5-√4=√5-2

√(9+4√5)=√(9+√80)=√[9+√(81-80)/2+√[9+√(81-80)/2=√(9+√1)/2+√(9-√1)/2=
=√(9+1)/2+√(9-1)/2=√5+√4=√5+2

√5-2+x(√5-2)=x(2+√5)-(√5+2)
√5-2+x√5-2x=2x+x√5-√5-2
√5=4x-√5
4x=2√5
x=√5/2



sau

√(2-√5)²+x√(9-4√5)=x√(2+√5)²-√(9+4√5)
√(2-√5)²+x√(4-4√5+5)=x√(2+√5)²-√(4+4√5+5)
√(2-√5)²+x√(2-√5)²=x√(2+√5)²-√(2+√5)²
√5-2+x(√5-2) =x(2+√5)-(2+√5)
√5-2+x√5-2x =2x+x√5-2-√5
 √5-2x =2x-√5
4x=2√5
x=2√5/4
x=√5/2