Question

Encontre a fracao geratriz de cada dizima a seguir a) 0, 333 b)0, 121212 c)1, 444 d)1, 3222 e)0, 31444 f)0, 666 g)1, 3111 h)10, 222 i)0, 1333

Answer 1

a) 0,333... = 3/9 = 1/3

b) 0,1212... = 12/99 = 4/33

c) 1,444.. = 1 4/9 = (9 x 1 + 4)/9 = 13/9

d) 1,32222... 1 (32 - 3)/90 = 1 (29/90) = (90 x 1 + 29)/90 = 119/90 

e) 0,31444... = (314 - 31)/900 = 283/900 

f) 0,666... = 6/9 = 2/3

g) 1,3111.. 1 (31 - 3)/90 = 1 28/90 = (90 x 1 + 28)/90 = 118/90

h) 10,222...  10 2/9 = (9 x 10 + 2)/9 = 92/9

i) 0,1333... = (13 - 1)/90 = 12/90 =4/30 = 2/15

Espero ter ajudado.

Answer 2

a)0,333..... = 3/9 = 1/3

b) 0,1212 ...= 12/99 = 4/33

c) 1.444 ...= 14/9 =(9 x 1 + 4) = 13/9

d) 1,3222....1 (32-3)/90 = 1 (29/90) = (90 x 1 + 29)/90 = 119/90

e) 0,31444... = (314-31)/900 = 283/900

f) 0,666... = 6/9 = 2/3

g) 1,3111...  1(31-3)/90 = 129/90 = (90 x 1 + 28/90) =118/90

h) 10,222 = 10(2/9) = (9 x 10 + 2)/90 = 92/90

i) 0,1333... =(13-1)/90 = 12/90= 4/30= 2/15