$A = \dfrac{x^2-2x+2018}{2017x^2}$ $ĐKXĐ : x \neq 0$
$ = \dfrac{1}{2017}.\bigg(1-\dfrac{2}{x}+\dfrac{2018}{x^2}\bigg)$
$ = \dfrac{1}{2017}. 2018.\bigg[\dfrac{1}{2018}-\dfrac{1}{x.1009}+\dfrac{1}{x^2}\bigg]$
$= \dfrac{1}{2017}.2018.\bigg[\bigg(\dfrac{1}{x}\bigg)^2-2.\dfrac{1}{x}.\dfrac{1}{2018}+\bigg(\dfrac{1}{2018}\bigg)^2\bigg]+\dfrac{2018-1}{2018^2}\bigg]$
$ = \dfrac{2018}{2017} . \bigg(\dfrac{1}{x}-\dfrac{1}{2018}\bigg)^2 + \dfrac{1}{2017}.\dfrac{2017}{2018}$
$ = \dfrac{2018}{2017} . \bigg(\dfrac{1}{x}-\dfrac{1}{2018}\bigg)^2 + \dfrac{1}{2018} ≥ \dfrac{1}{2018}$
Dấu "=" xảy ra $⇔ x=2018$
Vậy $A_{min} = \dfrac{1}{2018}$ tại $x=2018$
