Question

3) encontre a fração geratriz das dízimas periódicas simples a seguir:

a) 2,444...
b) 0,111...
c) 6,353535...
d) 2,102102102...

pfv me ajudem é uma questão de vida ou morte​

Answer 1

a) 2.444...

$x=2.444\dots\ \to\ x=2.\bar{4}\ \therefore\ 10x=24.\bar{4}\\\\ 10x-x=24.\bar{4}-2.\bar{4}\ \therefore\ 9x=22\ \therefore\ \boxed{x=\dfrac{22}{9}}$

b) 0.111...

$x=0.111\dots\ \to\ x=0.\bar{1}\ \therefore\ 10x=1.\bar{1}\\\\ 10x-x=1.\bar{1}-0.\bar{1}\ \therefore\ 9x=1\ \therefore\ \boxed{x=\dfrac{1}{9}}$

c) 6.353535...

$x=6.3535\dots\ \to\ x=6.\bar{35}\ \therefore\ 100x=635.\bar{35}\\\\ 100x-x=635.\bar{35}-6.\bar{35}\ \therefore\ 99x=629\ \therefore\ \boxed{x=\dfrac{629}{99}}$

d) 2.102102102...

$x=2.102102\dots\ \to\ x=2.\bar{102}\ \therefore\ 1000x=2102.\bar{102}\\\\ 1000x-x=2102.\bar{102}-2.\bar{102}\ \therefore\ 999x=2100\ \therefore\ \boxed{x=\dfrac{700}{333}}$

Answer 2

Explicação passo-a-passo:

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a)Temos a seguinte dízima periódica.

$x=\text{2,444...}$

Multiplicando a dízima por 10 e por 1:

$10x=\text{24,444...}\\x=\text{2,444...}$

Subtraindo:

$10x-x=\text{24,444...}-\text{2,444...}\\\\9x=22\\\\x=\dfrac{22}{9}\\\\\boxed{\boxed{\text{2,444...}=\dfrac{22}{9} }}$

b)Temos a seguinte dízima periódica.

$x=\text{0,111...}$

Multiplicando a dízima por 10 e por 1:

$10x=\text{1,111...}\\x=\text{0,111...}$

Subtraindo:

$10x-x=\text{1,111...}-\text{0,111...}\\\\9x=1\\\\x=\dfrac{1}{9}\\\\\boxed{\boxed{\text{0,111...}=\dfrac{1}{9} }}$

c)Temos a seguinte dízima periódica.

$x=\text{6,3535...}$

Multiplicando a dízima por 100 e por 1:

$100x=\text{635,3535...}\\x=\text{6,3535...}$

Subtraindo:

$100x-x=\text{635,3535...}-\text{6,3535...}\\\\99x=629\\\\x=\dfrac{629}{99} \\\\\boxed{\boxed{\text{6,3535...}=\dfrac{629}{99}}}$

d)Temos a seguinte dízima periódica.

$x=\text{2,10210...}$

Multiplicando a dízima por 1000 e por 1:

$1000x=\text{2102,10210...}\\x=\text{2,10210...}$

Subtraindo:

$1000x-x=\text{2102,10210...}-\text{2,10210...}\\\\999x=2100\\\\x=\dfrac{2100}{999}=\dfrac{700}{333} \\\\\boxed{\boxed{ \text{2,10210...}=\dfrac{700}{333}}}$