Question

Determine:
a) 0,5 + 0,777... - 0,9
b)$\frac{\sqrt{0,111...} }{0,3}$

Answer 1

Como temos dizimas periódicas, vamos precisar utilizar as frações geratrizes dos termos apresentados.

Por praticidade, vamos primeiro achar todas as geratrizes e, depois, podemos calcular o valor da expressões.

a)

     $\begin{array}{ccc}\boxed{\begin{array}{l}0,5~=\\\\\\=~0,5\cdot\dfrac{10}{10}\\\\\\=~\dfrac{5}{10}\\\\\\=~\boxed{\dfrac{1}{2}}\end{array}}&\boxed{\begin{array}{l}\underline{Seja~~x=0,777...}\\\\10x~=~10\cdot0,777...\\\\\boxed{10x~=~7,777...}\\\\\\10x-x~=~7,777...-0,777...\\\\9x~=~7\\\\\boxed{x~=~\dfrac{7}{9}}\end{array}}&\boxed{\begin{array}{l}0,9~=\\\\\\=~0,9\cdot\dfrac{10}{10}\\\\\\=~\boxed{\dfrac{9}{10}}\end{array}}\end{array}$

$0,5~+~0,777...~-~0,9~=\\\\\\=~\dfrac{1}{2}~+~\dfrac{7}{9}~-~\dfrac{9}{10}\\\\\\=~\dfrac{45\cdot1~+~10\cdot7~-~9\cdot9}{90}\\\\\\=~\dfrac{45~+~70~-~81}{90}\\\\\\=~\dfrac{34}{90}\\\\\\=~\boxed{\dfrac{17}{45}~ou~0,3777...}$

b)

$\begin{array}{ccc}\boxed{\begin{array}{l}\underline{Seja~~x=0,111...}\\\\10x~=~10\cdot0,111...\\\\\boxed{10x~=~1,111...}\\\\\\10x-x~=~1,111...-0,111...\\\\9x~=~1\\\\\boxed{x~=~\dfrac{1}{9}}\end{array}}&\boxed{\begin{array}{l}0,3~=\\\\\\=~0,3\cdot\dfrac{10}{10}\\\\\\=~\boxed{\dfrac{3}{10}}\end{array}}\end{array}$

$\dfrac{\sqrt{0,111...}}{0,3}~=\\\\\\=~\dfrac{\sqrt{\frac{1}{9}}}{\frac{3}{10}}\\\\\\=~\dfrac{\frac{1}{3}}{\frac{3}{10}}\\\\\\=~\dfrac{1}{3}~\cdot\dfrac{10}{3}\\\\\\=~\boxed{\dfrac{10}{9}~~ou~~1,111...}$