Question

Ma poate ajuta cineva la problema aceasta? :
Aflati trei numere consecutive invers proportionale cu $\frac{1}{2}$ , $\frac{2}{5}$ si $\frac{1}{3}$ .
Multumesc!

Answer 1

fie nr consecutive a, a+1, a+2 invers prop cu 1/2, 2/5 si 1/3
a*1/2=(a+1)*2/5=(a+2)*1/3=k
a=2k
a+1=5k/2
a+2=3k
a+(a+1)+(a+2)=3a+3=6k+3
2k+5k/2+3k=6k+3
4k+5k+6k=12k+6
15k-12k=6
3k=6
k=6:3=2
Nr sunt:
a=2*2=4
a+1=5
a+2=6