Question

Dica: o resultado é encontrado através das regras de uma dízima periódica

Answer 1

$\dfrac{1,101010...+0,1111...}{0,09696...} = \dfrac{ \dfrac{110-1}{99} + \dfrac{1}{9} }{ \dfrac{96}{990} } = \dfrac{ \dfrac{109}{99} + \dfrac{1}{9} }{ \dfrac{16}{165} } = \dfrac{ \dfrac{109+11}{99}}{ \dfrac{16}{165} } = \dfrac{ \dfrac{120}{99}}{ \dfrac{16}{165} }$$= \dfrac{ \dfrac{40}{33}}{ \dfrac{16}{165} } =\dfrac{40}{33}.\dfrac{165}{16}=\dfrac{10}{11}.\dfrac{55}{4}=\dfrac{5}{1}.\dfrac{5}{2}= \dfrac{25}{2} =12,5$