Question

A soma das dízimas periódicas infinitas 2,76666. E 1,4444. É: *

Answer 1

Resposta:

Denotando por a e b, nessa ordem, as dízimas dadas, temos:

$\sf a+b=2.7\overline{6}+1.\overline{4}$

$\sf a+b=(2+0.7\overline{6})+(1+0.\overline{4})$

$\sf a+b=3+(0.7\overline{6}+0.\overline{4})$

Denotando por c e d, nessa ordem, as dízimas 0,7666... e 0,444...:

$\sf c=0.7\overline{6}\implies 10c=7.\overline{6}~e~100c=76.\overline{6}$

$\sf\therefore 100c-10c=76.\overline{6}-7.\overline{6}\implies99c=69\implies c=\dfrac{69}{90}$

$\sf d=0.\overline{4}\implies 10c=4.\overline{4}$

$\sf\therefore 10d-d=4.\overline{4}-0.\overline{4}\implies9d=4\implies d=\dfrac{4}{9}$

Assim temos:

$\sf a+b=3+\dfrac{69}{90}+\dfrac{4}{9}$

$\sf a+b=\dfrac{270}{90}+\dfrac{69}{90}+\dfrac{40}{90}$

$\sf a+b=\dfrac{270+69+40}{90}$

$\red{\sf a+b=\dfrac{379}{90}}$