Question

determine o valor de y=
$\sqrt{0.111...} + \frac{14}{0.777...}$

Answer 1

$\sqrt{0,1111...} = \sqrt{x} \\\\x=0,1111...\\10x=1,111...\\10x=1+0,1111...\\10x=1+x\\9x=1\\\\x= \dfrac{1}{9}\\\\ \sqrt{0,1111...} = \sqrt{ \dfrac{1}{9} }= \dfrac{1}{3}$

$\dfrac{14}{0,777...}= \dfrac{14}{a}\\\\\\a=0,777...\\10a=7,777...\\10a=7+0,777...\\10a=7+a\\9a=7\\\\a= \dfrac{7}{9}\\\\\\ \dfrac{14}{0,777...} = \frac{14}{ \dfrac{7}{9} }\\\\\\ \dfrac{14}{0,777...}=14\cdot \dfrac{9}{7}\\\\\\ \dfrac{14}{0,777...}=18$

$\sqrt{0,1111...}+ \dfrac{14}{0,777...}= \dfrac{1}{3}+18\\\\\\ \sqrt{0,1111...}+ \dfrac{14}{0,777...}= \dfrac{1+54}{3}\\\\\\\boxed{y= \sqrt{0,1111...}+ \dfrac{14}{0,777...}= \frac{55}{3}}$