Question

0,0999... é uma dízima periódica. Qual é a geratriz dessa
dízima?

Answer 1

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$\Large\boxed{\begin{array}{l}\sf x=0,0999...\cdot100\\\sf 100x=9,999....\cdot10\\\sf 1000x=99,999...\\-\underline{\begin{cases}\sf 1000x=99,999...\\\sf 100x=9,999...\end{cases}}\\\sf 900x=90\\\sf x=\dfrac{90\div90}{900\div90}\\\sf x=\dfrac{1}{10}\end{array}}$

$\boxed{\begin{array}{l}\underline{\rm pela~soma~dos~termos~da~PG~infinita\!:}\\\sf 0,0999...=\underbrace{0,09+0,009+0,0009+...}_{\sf soma~dos~termos~da~PG~infinita.}\\\sf a_1=0,09=\frac{9}{100}\\\sf a_2=0,009=\frac{9}{1000}\\\sf q=\dfrac{a_2}{a_1}\\\sf q=\dfrac{\frac{9}{1000}}{\frac{9}{100}}\\\sf q=\dfrac{\diagdown\!\!\!9}{10\diagdown\!\!\!\!0\diagdown\!\!\!\!0}\cdot\dfrac{1\diagdown\!\!\!0\diagdown\!\!\!0}{\diagdown\!\!\!9}=\dfrac{1}{10}\end{array}}$

$\boxed{\begin{array}{l}\sf S_n=\dfrac{a_1}{1-q}\\\sf S_n=\dfrac{\frac{9}{100}}{1-\frac{1}{10}}\\\sf S_n=\dfrac{\frac{9}{100}}{\frac{9}{10}}\\\sf S_n=\dfrac{\diagup\!\!\!9}{10\diagdown\!\!\!\!0}\cdot\dfrac{1\diagdown\!\!\!\!0}{\diagup\!\!\!9}=\dfrac{1}{10}\end{array}}$