Giải : Ta có: A=\(\dfrac{x^2-2x+2017}{x^2}\)
=\(\dfrac{\left(x-1\right)^2}{x^2}+\dfrac{2016}{x^2}\)\(\ge\) \(\dfrac{2016}{x^2}\)
Dấu đẳng thức xảy ra \(\Leftrightarrow\) \(\left(x-1\right)^2=0\Leftrightarrow x=1\)
Vậy Min A = 2016 \(\Leftrightarrow\) x=1