rezolvarea primei probleme este urmatoarea:( 2^4=2 la puterea a 4-a)
2√18(√32-√2)=2√(2*3^2)*(√(2*2^4)-√2)=
=2*3V2*(2²*√2-√2)=
=2*3√2*√2(2²-1)=
=2*3*2*3=
=36
A doua problema:
20/√20+30/√180-35/√245=
20/√(2²*5)+30/V(3²*2²*5)-35/√(7²*5)=
20/2√5+30/(3*2*√5)-35/(7√5)=
10/√5+5/√5-5/√5=
(10+5-5)/√5=10/√5=(10√5)/5=2√5
