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a fost răspuns

fractia ordinara ireductibila rezultata din fractia 3+6+9+12+.....+150 supra 5+10+15+20+.....+250 prin simplificarea acesteia este?

Răspuns :

DeeeaFlorina
3+6+9+12+...+150=
=3×(1+2+3+4+...+50)
=3×50×(50+1):2
=3×50×51:2

5+10+15+20+...+250=
=5×(1+2+3+4+...+50)
=5×50×(50+1):2
=5×50×51:2

3×50×51:2 supra 5×50×51:2
(Simplificam cu 50×51:2)
=3 supra 5
3+6+9+12+.....+150 supra 5+10+15+20+.....+250


(3+6+9+12+.....+150)/(5+10+15+20+.....+250)


[tex]\it \dfrac{(3+6+9+12+.....+150)}{(5+10+15+20+.....+250} = \dfrac{3(1+2+3+ ... +50)}{5(1+2+3+ ... +50)} = \dfrac{3}{5}[/tex]

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