ĐKXĐ: \(x\ne0\)
\(\dfrac{x^2-2x+2007}{2007x^2}=\dfrac{2007x^2-2.2007x+2007^2}{2007^2.x^2}\)\(\Rightarrow\dfrac{\left(x-2007\right)^2}{2007^2.x^2}+\dfrac{2006}{2007^2}\ge\dfrac{2006}{2007^2}\)
Dấu " = " xảy ra \(\Leftrightarrow x=2007\)
Vậy min = \(\dfrac{2006}{2007^2}\)