Question

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calcule a fração geratriz das seguintes dizimas periódicas, usando o limite da soma de uma p.g infinita

a) 3,888888...

b) 2,5555...​

Answer 1

Explicação passo-a-passo:

a) $\sf 3,888\dots=3+0,8+0,08+0,008+\dots$

$\sf q=\dfrac{0,08}{0,8}$

$\sf q=\dfrac{8}{80}$

$\sf q=\dfrac{1}{10}$

$\sf S=\dfrac{a_1}{1-q}$

$\sf S=\dfrac{0,8}{1-\frac{1}{10}}$

$\sf S=\dfrac{\frac{8}{10}}{\frac{9}{10}}$

$\sf S=\dfrac{8}{9}$

Logo:

$\sf 3,888\dots=3+\dfrac{8}{9}$

$\sf 3,888\dots=\dfrac{3\cdot9+8}{9}$

$\sf 3,888\dots=\dfrac{27+8}{9}$

$\sf 3,888\dots=\dfrac{35}{9}$

b) $\sf 2,555\dots=2+0,5+0,05+0,005+\dots$

$\sf q=\dfrac{0,05}{0,5}$

$\sf q=\dfrac{5}{50}$

$\sf q=\dfrac{1}{10}$

$\sf S=\dfrac{a_1}{1-q}$

$\sf S=\dfrac{0,5}{1-\frac{1}{10}}$

$\sf S=\dfrac{\frac{5}{10}}{\frac{9}{10}}$

$\sf S=\dfrac{5}{9}$

Logo:

$\sf 2,555\dots=2+\dfrac{5}{9}$

$\sf 2,555\dots=\dfrac{2\cdot9+5}{9}$

$\sf 2,555\dots=\dfrac{18+5}{9}$

$\sf 2,555\dots=\dfrac{23}{9}$