Easy example that shows [tex]0.999\ldots=1[/tex]:
Let [tex]x=0.999\ldots=0.\overline9[/tex].
Then [tex]10x=9.999\ldots=9.\overline9[/tex].
So [tex]10x-x=9.\overline9-0.\overline9[/tex], or [tex]9x=9[/tex], and so [tex]x=1[/tex].
The basic idea is to find the period of the repeating decimal, move the [tex]n[/tex] digits belonging to one period over to the left of the decimal point by multiplying by [tex]10^n[/tex], then subtract the original repeating decimal from this new number, and finally divide by [tex]10^n-1[/tex].
A slightly more complicated example:
Let [tex]x=0.142857142857142857\ldots=0.\overline{142857}[/tex].
Then [tex]10^6x=142857.\overline{142857}[/tex].
Then [tex]10^6x-x=142857.\overline{142857}-0.\overline{142857}[/tex], or
[tex]999999x=142857\implies x=\dfrac{142857}{999999}=\dfrac17[/tex]
