Question

Determine a fração geratriz das dízimas peródicas abaixo b)0,2444...​

Answer 1

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• vou resolver de duas formas:

1. pela soma dos termos da  PG infinita
2. pela equação de 1º grau

$\underline{\rm pela~soma~dos~termos~da~PG~infinita\!:}$

$\boxed{\begin{array}{l}\sf 0,2444...=0,2+\underbrace{0,04....+0,004...+0,0004}_{ soma~dos~termos~da~PG~infinita}\\\sf a_1=0,04=\dfrac{4}{100}~~a_2=\dfrac{4}{1000}\\\sf q=\dfrac{a_2}{a_1}\\\sf q=\dfrac{\frac{4}{1000}}{\frac{4}{100}}\\\sf q=\dfrac{\backslash\!\!\!4}{10\diagdown\!\!\!\!\!\!00}\cdot\dfrac{1\diagdown\!\!\!\!\!\!00}{\backslash\!\!\!4}=\dfrac{1}{10}\end{array}}$

$\boxed{\begin{array}{l}\sf S_n=\dfrac{a_1}{1-q}\\\sf S_n=\dfrac{\frac{4}{100}}{1-\frac{1}{10}}\\\sf S_n=\dfrac{\frac{4}{100}}{\frac{9}{10}}\\\sf S_n=\dfrac{4}{10\backslash\!\!\!0}\cdot\dfrac{1\backslash\!\!\!0}{9}\\\sf S_n=\dfrac{4}{90}\\\sf 0,2444...=0,2+\dfrac{4}{90}\\\sf 0,2444...=\dfrac{2}{10}+\dfrac{4}{90}\\\sf 0,2444...=\dfrac{18+4}{90}\\\sf 0,2444...=\dfrac{22\div2}{90\div2}\\\\\sf 0,2444...=\dfrac{11}{45}\end{array}}$

$\underline{\rm pela~equac_{\!\!,}\tilde ao\!:}$

$\boxed{\begin{array}{l}\sf x=0,2444...\cdot10\\\sf 10x=2,444....\cdot10\\\sf 100x=24,444....\\-\underline{\begin{cases}\sf 100x=24,444....\\\sf 10x=2,444....\end{cases}}\\\sf 90x=22\\\sf x=\dfrac{22\div2}{90\div2}\\\\\sf x=\dfrac{11}{45}\end{array}}$