A=\(1-\frac{2}{x}+\frac{2014}{x^2}=1-\frac{2.\sqrt{2014}}{x}.\frac{1}{\sqrt{2014}}+\left(\frac{\sqrt{2014}}{x}\right)^2=\left(\frac{\sqrt{2014}}{x}\right)^2-\frac{2}{x}+\frac{1}{2014}+\frac{2013}{2014}=\left(\frac{\sqrt{2014}}{x}-\frac{1}{\sqrt{2014}}\right)^2+\frac{2013}{2014}\ge\frac{2013}{2014}\)
Vậy Min A là 2013/2014 với x=2014