• Matéria: Matemática
  • Autor: micaelsuzuki
  • Perguntado 9 anos atrás

Resolva : a- 0,555... b-2,363636... c-1,0909090... d-5,018018180... e-1,04727272 f-1,32444 g-1,05333 h-1,0303030 i-0,0666... j-2,027027027 k-0,003003003... l-3,61666... m-2,068181181... n-1,291666...

Respostas

respondido por: exalunosp
3
0,555..... 5/9
b
2,36 36.....  2 inteiros  36/99  = 2 int 4/11 ( 11*2)+4 = 26/11 ***
c
1,09 09 09..... 1 int 9/99 = 1 int 1/11  ( 11*1)+1 = 12/99***
d
5, 018 018 018... = 5 int  18/999 = 5 int  2/111 = ( 111 * 2) + 5 = 118/111
e
1,04 72 72 72 = 1 int ( 0472 - 04)/9900 = 1 int 468/9900 =  1 int  13/275=     = ( 275 *1)+13 = 288/275

1,32 444.... = 1 int  ( 324 - 32 )/900 = 1 int 292/900 = 1 int   73/225 =            ( 225 * 1) + 73 = 298/225
g
1,05 333.... = 1 int  (053 - 05 )/900  = 1 int  48/900 = 1 int   12/225=             1 int 4/75 = ( 75 * 1)+4 = 79/75
h
1,03 03 03.....  = 1 int 03/99  = 1 int  1/33 = ( 33*1) + 1 =34/33
i
0,0 666..... = 6/90 = 1/15
j
2, 027 027 ..... = 2 int 027/999 = 2 int  3/111 = ( 111 * 2) + 3 = 225/111
k
0,003 003 003..... =  3/999 = 1/333 ****
l
3,61 666..... = 3 int  ( 616 - 61)900 = 3 int 555/900  = 3 int 111/180 =            = ( 180 * 3 ) + 111 = 651/180 = 217/60
 m
 2, 068 181 181....= 2 int ( 068181-068)/999000 = 2 int  68113/999000 =      ( 999000 * 2) + 68113 =  2066113/999000 
n
 1,291 666..... = 1 int  ( 2916 -291 )/9000  = 1 int 2625/9000 =                        1 int  105/360 =  1 int 21/72  ( 72 * 1) + 21 = 93/72 

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