Ta có: \(\left(x+\sqrt{x^2+5}\right)\left(y+\sqrt{y^2}+5\right)=5\\ \Leftrightarrow\left(x+\sqrt{x^2+5}\right)\left(\sqrt{x^2+5}-x\right)\left(y+\sqrt{y^2+5}\right)=5\left(\sqrt{x^2+5}-x\right)\\ \Leftrightarrow5\left(y+\sqrt{y^2+5}\right)=5\left(\sqrt{x^2+5}-x\right)\\ \Leftrightarrow x+y=\sqrt{x^2+5}-\sqrt{y^2+5}\left(1\right)\)Mặt khác:
\(\left(x+\sqrt{x^2+5}\right)\left(y+\sqrt{y^2+5}\right)=5\\ \Leftrightarrow\left(x+\sqrt{x^2+5}\right)\left(\sqrt{y^2+5}-y\right)\left(y+\sqrt{y^2+5}\right)=5\left(\sqrt{y^2+5}-y\right)\\ \Leftrightarrow5\left(x+\sqrt{x^2+5}\right)=5\left(\sqrt{y^2+5}-y\right)\\ \Leftrightarrow x+y=\sqrt{y^2+5}-\sqrt{x^2+5}\left(2\right)\)Cộng (1) và (2) theo vế ta có: \(x+y=0\)
