Question

Stii d ca a si b sunt cifre nenule, calculati..... ​

Answer 1

 

$\displaystyle\bf\\b)\\\left(\frac{1}{a}+\frac{1}{\overline{aa}}+\frac{1}{\overline{aaa}}\right):\left(\frac{1}{b}+\frac{1}{\overline{bb}}+\frac{1}{\overline{bbb}}\right)=\\\\\\=\frac{\left(\dfrac{1}{a}+\dfrac{1}{\overline{aa}}+\dfrac{1}{\overline{aaa}}\right)}{\left(\dfrac{1}{b}+\dfrac{1}{\overline{bb}}+\dfrac{1}{\overline{bbb}}\right)}=$

.

$\displaystyle\bf\\=\frac{\left(\dfrac{1}{a}+\dfrac{1}{11a}+\dfrac{1}{111a}\right)}{\left(\dfrac{1}{b}+\dfrac{1}{11b}+\dfrac{1}{111b}\right)}=\\\\\\=\frac{\dfrac{1}{a}\left(1+\dfrac{1}{11}+\dfrac{1}{111}\right)}{\dfrac{1}{b}\left(1+\dfrac{1}{11}+\dfrac{1}{111}\right)}=\\\\\\=\frac{~~\dfrac{1}{a}~~}{\dfrac{1}{b}}=\dfrac{1}{a}\times\dfrac{b}{1}=\boxed{\bf\dfrac{b}{a}}$