Đặt \(A=\)\(\frac{x^2-2x+2016}{x^2}\)(ĐKXĐ: x≠0)
\(\Leftrightarrow x^2\left(A-1\right)+2x-2016=0\)
△ = \(2^2+4.2016\left(A-1\right)\ge0\)
\(\Leftrightarrow8064A\ge8060\Leftrightarrow A\ge\frac{2015}{2016}\)
Vậy min A = \(\frac{2015}{2016}\Leftrightarrow\)\(x=2016\)