Đáp án: $Min A=\dfrac{1}{2018}$
Giải thích các bước giải:
Ta có :
$A=\dfrac{x^2-2x+2018}{2017x^2}$
$\to A=\dfrac{1}{2017}(1-\dfrac2x+\dfrac{2018}{x^2})$
$\to A=\dfrac1{2017}\left(2018\left(\dfrac1x-\dfrac{1}{2018}\right)^2+\dfrac{2017}{2018}\right)$
$\to A\ge \dfrac1{2017}.\dfrac{2017}{2018}$
$\to A\ge \dfrac1{2018}$
Dấu = xảy ra khi $x=2018$