Question

fração geratriz das dizimas o,424242 ; 2,383838 ; 2,141414 

Answer 1

$x = 0,424242...$

Multiplicando a equação por 100:

$100*x=100*0,424242...\\100x=42,424242...\\100x=42+0,424242...$

Como 0,424242... = x:

$100x=42+x\\100x-x=42\\99x=42\\x=42/99\\x=14/33$
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$2,383838... = 2 + 0,383838...$

$x=0,383838...\\100x=38,383838...\\100x=38+0,383838...\\100x=38+x\\99x=38\\x=38/99$

$2,383838...=2+0,383838...\\2,383838...=2+(38/99)\\2,383838...=(2*99/99)+(38/99)\\2,383838...=(198/99)+(38/99)\\2,383838...=(198+38)/99\\2,383838...=236/99$
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$2,141414... = 2+0,141414...$

Vou resolver essa pelos conceitos de P.G (progressão geométrica):

$0,141414... = 0,14 + 0,0014 + 0,000014 + ...$

$a_{1}=0,14=14/100\\a_{2}=0,0014=14/10000$

$q = a_{2}/a_{1}=(14/10000)/(14/100)=(14/10000)*(100/14)=1/100$

0 < q < 1: Soma dos termos infinitos de uma P.G

$S_{n}=a_{1}/(1-q)\\0,141414...=a_{1}/(1-q)\\0,141414...=(14/100)/(1-[1/100])\\0,141414...=(14/100)/([100-1]/100)\\0,141414...=(14/100)/(99/100)\\0,141414...=(14/100)*(99/100)\\0,141414...=14/99$

$2,141414... = 2+0,141414...\\2,141414...=2+(14/99)\\2,141414...=(2*99/99)+(14/99)\\2,141414...=(198/99)+(14/99)\\2,141414...=(198+14)/99\\2,141414...=212/99$

Answer 2

0,42424242 = 42/99
x=0,424242
100x=42,424242
100x-x=42.42-0,42
99x=42
x=42/99

2,3838383 = 236/99
x=2,383838
100x=238,383838
100x-x=238,38-2,38
99x=236
x=236/99

2,141414 = 212/99
x=2,141414
100x=214,14
100x-x = 214,14-2,14
99x=212
x=212/99