Question

Encontre a fração geratriz das dízimas periódicas simples a seguir:
a) 0,33333... f) 0,44444...

b) 0,11111.. g) 0,6666...

c) 17,88888...
d) -6,353535....
e) 0,292929...

Answer 1

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$\tt{a)}~\sf{x=0,333...\cdot10}\\\sf{10x=3,333...}\\-\underline{\begin{cases}\sf{10x=3,333...}\\\sf{x=0,333...}\end{cases}}\\\sf{9x=3}\\\sf{x=\dfrac{3}{9}}\\\huge\boxed{\boxed{\boxed{\boxed{\sf{x=\dfrac{1}{3}}}}}}$

$\tt{b)}~\sf{x=0,111....\cdot10}\\\sf{10x=1,111...}\\-\underline{\begin{cases}\sf{10x=1,111...}\\\sf{x=0,111...}\end{cases}}\\\sf{9x=1}\\\huge\boxed{\boxed{\boxed{\boxed{\sf{x=\dfrac{1}{9}}}}}}$

$\tt{c)}~\sf{k=17,888...\cdot10}\\\sf{10k=178,888...}\\-\underline{\begin{cases}\sf{10k=178,888...}\\\sf{k=17,888...}\end{cases}}\\\sf{9k=161}\\\\\huge\boxed{\boxed{\boxed{\boxed{\boxed{\sf{k=\dfrac{161}{9}}}}}}}$

$\tt{d)}~\sf{m=6,353535...\cdot100}\\\sf{100m=635,3535....}\\-\underline{\begin{cases}\sf{100m=635,3535...}\\\sf{m=6,3535...}\end{cases}}\\\sf{99m=629}\\\sf{m=\dfrac{629}{99}}\\\boxed{\boxed{\boxed{\boxed{\sf{-6,3535...=-\dfrac{629}{99}}}}}}$

$\tt{e)}~\sf{n=0,2929...\cdot100}\\\sf{100n=29,2929...}\\-\underline{\begin{cases}\sf{100n=29,2929...}\\\sf{n=0,2929...}\end{cases}}\\\sf{99n=29}\\\ \huge\boxed{\boxed{\boxed{\boxed{\sf{n=\dfrac{29}{99}}}}}}$

$\tt{f)}~\sf{0,444...=4\cdot0,111...=4\cdot\dfrac{1}{9}=\dfrac{4}{9}}$

$\tt{g)}~\sf{0,666...=2\cdot0,333...=2\cdot\dfrac{1}{3}=\dfrac{2}{3}}$