Đặt \(t=\frac{1}{x+2010}\Rightarrow x=\frac{1}{t}-2010\)
Ta có: \(E=x\cdot\frac{1}{\left(x+2010\right)^2}=\left(\frac{1}{t}-2010\right)t^2=t-2010t^2\)
\(=-2010\left(t^2-t\cdot\frac{1}{2010}\right)=-2010\left(t^2-2t\cdot\frac{1}{4020}+\frac{1}{4020^2}\right)+\frac{1}{8040}\)
\(=-2010\left(t-\frac{1}{4020}\right)^2+\frac{1}{8040}\le\frac{1}{8040}\)
Dấu "=" xảy ra <=> \(t=\frac{1}{4020}\Leftrightarrow x=2010\)