Question

24. Rationalizati suma inverselor numerelor a, b, c unde:

Answer 1

1/a = 1/(3rad2-2rad3) = (3rad2+2rad3)/(18-12) = (3rad2+2rad3)/6
1/b = 1/(3-2rad2) = (3+2rad2)/(9-8) = 3+2rad2
1/c = 1/(3rad3-5) = (3rad3+5)/(27-25) = (3rad3+5)/2

1/a + 1/b + 1/c = (3rad2+2rad3)/6 + (3+2rad2) + (3rad3+5)/2 =
Numitor comun 6
(3rad2+2rad3+18+12rad2+9rad3+15)/6 =
(15rad2+11rad3+33)/6