Ta có\(\left(x+\sqrt{x^2+2012}\right)\left(y+\sqrt{y^2+2012}\right)=2012\) đk \(\left\{{}\begin{matrix}x^2+2012\ge0\\y^2+2012\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left(x-\sqrt{x^2+2012}\right)\left(x+\sqrt{x^2+2012}\right)\left(y+\sqrt{y^2+2012}\right)=2012\left(x-\sqrt{x^2+2012}\right)\)
\(\Leftrightarrow\left(x^2-x^2-2012\right)\left(y+\sqrt{y^2+2012}\right)=2012\left(x-\sqrt{x^2+2012}\right)\)
\(\Leftrightarrow-2012\left(y+\sqrt{y^2+2012}\right)=2012\left(x-\sqrt{x^2+2012}\right)\)
\(\Leftrightarrow-y-\sqrt{y^2+2012}=x-\sqrt{x^2+2012}\)
\(\Leftrightarrow x+y-\left(\sqrt{x^2+2012}-\sqrt{y^2+2012}\right)=0\)
\(\Leftrightarrow x+y-\dfrac{\left(x-y\right)\left(x+y\right)}{\sqrt{x^2+2012}+\sqrt{y^2+2012}}=0\)
\(\Leftrightarrow\left(x+y\right)\left(1-\dfrac{x-y}{\sqrt{x^2+2012}+\sqrt{y^2+2012}}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\1=\dfrac{x-y}{\sqrt{x^2+2012}+\sqrt{y^2+2012}}\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow E=0\)