Question

Determine a fraçao geratriz
A) 0,2111...
B) 2,141414...
C) 0,323232...
D) 3,2111...
E) 0,321321321...
F) 4,121212...

Answer 1

a) 0,21111...

y = 0,21111...
10y = 2,11111....

2,1111... = 2 + 0,11111...

x = 0,11111...
10x = 1,1111...
10x = 1 + 0,1111...
10x = 1 + x
10x - x = 1
9x = 1
x = 1/9

2,1111.... = 2 + 0,1111...
2,1111... = 2 + 1/9 = (18+1) / 9 = 19/9

10y = 2,11111....
10y = 19/9

y = 19/90

b) 2,141414.... = 2 + 0,141414...

x = 0,141414...

100x = 14,1414...
100x = 14 + 0,141414...
100x = 14 + x
100x - x = 14
99x = 14
x = 14/99

2,14141414... = 2 + 0,141414... = 2 + 14/99 = (198+14) / 99 = 212/99

c) 0,32323232...

x = 0,32323232...
100x = 32,323232...
100x = 32 + 0,323232...
100x = 32 + x
100x - x = 32
99x = 32
x = 32/99

d) 3,2111...

3,211111... = 3 + 0,211111...
y = 0,21111...
10y = 2,11111....

2,1111... = 2 + 0,11111...

x = 0,11111...
10x = 1,1111...
10x = 1 + 0,1111...
10x = 1 + x
10x - x = 1
9x = 1
x = 1/9

2,1111.... = 2 + 0,1111...
2,1111... = 2 + 1/9 = (18+1) / 9 = 19/9

10y = 2,11111....
10y = 19/9

y = 19/90

3,21111... = 3 + 0,21111.... = 3 + 19/90 = (270+19) / 90 = 289/90

e) 0,321321321...

x = 0,321321321...
1000x = 321 + 0,321321...
1000x = 321 + x
1000x - x = 321
999x = 321
x = 321/999

f) 4,121212... = 4 + 0,121212...

x = 0,121212...
100x = 12,1212...
100x = 12 + 0,121212...
100x = 12 +x
100x - x = 12
99x = 12
x = 12/99

4,121212... = 4 + 0,121212... = 4 + 12/99 = (396 + 12) / 99 = 408/99