Ta có:
+/\(\left(x+\sqrt{x^2+3}\right)\left(y+\sqrt{y^2+3}\right)=3\)
Mà \(\left(x+\sqrt{x^2+3}\right)\left(\sqrt{x^2+3}-x\right)=x^2+3-x^2=3\)
\(\Rightarrow y+\sqrt{y^2+3}=\sqrt{x^2+3}-x\)
\(\Leftrightarrow x+y=\sqrt{x^2+3}-\sqrt{y^2+3}\) (1)
+/\(\left(x+\sqrt{x^2+3}\right)\left(y+\sqrt{y^2+3}\right)=3\)
Mà \(\left(\sqrt{y^2+3}-y\right)\left(y+\sqrt{y^2+3}\right)=y^2+3-y^2=3\)
\(\Rightarrow x+\sqrt{x^2+3}=\sqrt{y^2+3}-y\)
\(\Leftrightarrow x+y=\sqrt{y^2+3}-\sqrt{x^2+3}\) (2)
Từ (1), (2)
\(\Rightarrow2\left(x+y\right)=0\)
\(\Leftrightarrow x+y=0\)
Vậy \(x+y=0\)