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