Question

Suma a 3 numere naturale este 2011.primul numarul este de 6 ori mai mic decit al doilea,iar al treilea numar este cu 173 mai mic decit primul.sa se afle cele trei numere.

Answer 1

a+b+c=2011 a=b/6 ; b=6a c=a-173 ; inlocuim pe b si c in prima relatie a +6a +a-173 =2011 8a =2011+173 8a =2184 a =273 b =6a b =6*273 =1638 c=a -173 c =273-173 =100 ( V : 273+1638+100 =2011 )

Answer 2

∅ Rezolvare :


_________________
Datele problemei :
_________________

___________________________________________________

→ a + b + c = 2 011
___________________________________________________

→ b = 6a
___________________________________________________

→ c = a — 173
___________________________________________________


→ a + 6a + a — 173 = 2 011 =>

→ 8a = 2 011 + 173 =>

→ 8a = 2 184 =>

→ a = 273
___________



→ b = 6a =>

→ b = 6 • 273 =>

→ b = 1 638
___________


→ c = a — 173 =>

→ c = 273 — 173 =>

→ c = 100
_____________