Viết lại đề : \(\left(x+\sqrt{x^2+2017}\right)\left(y+\sqrt{y^2+2017}\right)\)=2017 ( ? )
ta có: \(\left(\sqrt{x^2+2017}+x\right)\left(\sqrt{x^2+2017}-x\right)=x^2+2017-x^2=2017\)
mà \(\left(\sqrt{x^2+2017}+x\right)\left(\sqrt{y^2+2017}+y\right)=2017\)
\(\Rightarrow\sqrt{x^2+2017}-x=\sqrt{y^2+2017}+y\)
\(\Leftrightarrow x+y=\sqrt{x^2+2017}-\sqrt{y^2+2017}\)(1)
\(\left(\sqrt{y^2+2017}+y\right)\left(\sqrt{y^2+2017}-y\right)=2017\)
\(\Rightarrow\sqrt{y^2+2017}-y=\sqrt{x^2+2017}+x\)
\(\Leftrightarrow x+y=\sqrt{y^2+2017}-\sqrt{x^2+2017}\)(2)
giờ cộng vế với vế (1) và (2) ta có: \(2\left(x+y\right)=0\Leftrightarrow x+y=0\)