👤
Cristiansmc77
a fost răspuns

Fracție ordinara ireductibila 3+6+9+12+...+150/5+10+15+20+...250 prin simplificare acesteia este...

Răspuns :

3+6+9+12+....+150 / 5+10+15+20+.....+250=
=3*(1+2+3+4+.....+50) / 5*(1+2+3+4+.....+50)=3/5
[tex] \frac{3+6+9+12+...+150}{5+10+15+20+...250} =\ \textgreater \ \frac{3(1+2+3+...+50)}{5(1+2+3+...+50)} ,unde (1+2+3+...+50) se\\ simplifica=\ \textgreater \ \frac{3}{5} care~este~ireductibila~deoarece~nu~se~mai\\ poate~simplifica.[/tex]