\(\left(x+\sqrt{x^2+2019}\right)\left(\sqrt{x^2+2019}-x\right)=x^2+2019-x^2=2019\)
\(\Rightarrow\sqrt{x^2+2019}-x=y+\sqrt{y^2+2019}\left(2\right)\)
Tương tự \(\sqrt{y^2+2019}-y=x+\sqrt{x^2+2019}\left(1\right)\)
Lấy (2) - (1) được: -2x = 2y
<=> -x = y
<=> x + y = 0