Question

Suma a3 nr naturale eate 805.primele doua sunt direct proportionale 5 si 7 ,iar ultimele doua nr au raportul 9.
Care sunt numerele???
Coronita :)

Answer 1

a+b+c=805
b/c=9⇒b=9c⇒c=b/9
a+b+b/9=805
(9a+9b+b)/9=805
9a+10b=805x9
9a+10b=7245

{a,b}dir,prop{5,7}
a/5=b/7=k
a/5=k⇒a=5k
b/7=k⇒b=7k
inlocuim in 9a+10b=7245
9x5k+10x7k=7245
45k+70k=7245
115k=7245
k=7245/115
k=63

a=5k=5x63
a=315
b=7k=7x63
b=441
c=b/9=441/9
c=49

verificare
a+b+c=805
315+441+49=805