a/1/3 = 3 a
b/ 1/5 = 5 b
c/1/9 = 9 c
d/1/15 = 15 d
3 a = 5 b = 9 c = 15 d = k ⇒ a = k/3; b = k/5; c = k/9; d = k/15
2 k/3 - 3 k/5 + k/9 - k/15 = 15
cmmmc al numitorilor ( 3, 5, 9, 15 ) = 3²×5= 45
→ aducem fractiile la acelasi numitor comun 45
2 k ×15 - 3 k × 9 + k ×5 - k ×3 = 45 ×15
30 k - 27 k + 5 k - 3 k = 675
5 k = 675 ⇒ k = 675 : 5 ⇒ k = 135
a = 135/3 = 45
b = 135/5 = 27
c = 135/9 = 15
d = 135/15 = 9
Verific: 2×45 - 3 × 27 + 15 - 9 = 90 - 81 + 15 - 9 = 9 + 15 - 9 = 15
