Ta có \(\left(x+\sqrt{x^2+2017}\right).\left(y+\sqrt{y^2+2017}\right)=2017\)
\(\Rightarrow\frac{x^2-x^2-2017}{x-\sqrt{x^2+2017}}.\frac{y^2-y^2-2017}{y-\sqrt{y^2+2017}}=2017\)
\(\Leftrightarrow\left(x-\sqrt{x^2+2017}\right)\left(y-\sqrt{y^2+2017}\right)=2017\)
\(\Rightarrow\left(x+\sqrt{x^2+2017}\right)\left(y+\sqrt{y^2+2017}\right)=\left(x-\sqrt{x^2+2017}\right)\left(y-\sqrt{y^2+2017}\right)\)
\(\Leftrightarrow x\sqrt{y^2+2017}+y\sqrt{x^2+2017}=-x\sqrt{y^2+2017}-y\sqrt{x^2+2017}\)
\(\Leftrightarrow2x\sqrt{y^2+2017}=-2y\sqrt{x^2+2017}\)
\(\Leftrightarrow\hept{\begin{cases}x\ge0;y\le0\\4x^2\left(y^2+2017\right)=4y^2\left(x^2+2017\right)\end{cases}}\Leftrightarrow x=-y\)
Vậy \(x+y=0\)
