Question

Encontre a fração geratriz das dízimas periódicas : a)2,0909... b)3,151515...c)6,0666... d)0,5727272...

Answer 1

A) 2,0909... = 2  09/99 = (99 X 2 + 9)/99 = 207/99 = 69/33 = 23/11

B) 3,151515... = 3 15/99 = (99 x 3 + 15)/99 = 312/99 = 104/33

C) 6,06666... = 6 06-0/90 = (90 x 6 + 6)/90 =  546/90 = 273/45 = 91/15

D) 0,572727... = (5727 - 57)/9900 = 5670/9900

Espero ter ajudado.

Answer 2

Explicação passo-a-passo:

A) 2,0909... = 2 09/99 = (99 X 2 + 9)/99 = 207/99 = 69/33 = 23/11

B) 3,151515... = 3 15/99 = (99 x 3 + 15)/99 = 312/99 = 104/33

C) 6,06666... = 6 06-0/90 = (90 x 6 + 6)/90 = 546/90 = 273/45 = 91/15

D) 0,572727... = (5727 - 57)/9900 = 5670/9900