Ta có: \(\left(x+\sqrt{x^2+3}\right)\left(\sqrt{x^2+3}-x\right)=3\)
\(\left(y+\sqrt{y^2+3}\right)\left(\sqrt{y^2+3}-y\right)=3\)
Kết hợp với giả thiết ta có:
\(\sqrt{x^2+3}-x=y+\sqrt{y^2+3}\)
\(\sqrt{y^2+3}-y=x+\sqrt{x^2+3}\)
Cộng theo vế ta được: \(-\left(x+y\right)=x+y\)
\(\Rightarrow\)\(E=x+y=0\)
\(\left(x+\sqrt{x^2+3}\right)\left(y+\sqrt{y^2+3}\right)=3\)
\(\Leftrightarrow\left(x+\sqrt{x^2+3}\right)\left(x-\sqrt{x^2+3}\right)\left(y+\sqrt{y^2+3}\right)=3\left(x-\sqrt{x^2+3}\right)\)
\(\Leftrightarrow\left(x^2-x^2-3\right)\left(y+\sqrt{y^2+3}\right)=3\left(x-\sqrt{x^2+3}\right)\)
\(\Leftrightarrow-3\left(y+\sqrt{y^2+3}\right)=3\left(x-\sqrt{x^2+3}\right)\)
\(\Leftrightarrow-y-\sqrt{y^2+3}=x-\sqrt{x^2+3}\)(*)
Tương tự, nhân mỗi vế vs \(y-\sqrt{y^2+3}\), ta được:
\(-x-\sqrt{x^2+3}=y-\sqrt{y^2+3}\)(**)
Cộng (*) và (**) suy ra :
\(-y-x-\sqrt{y^2+3}-\sqrt{x^2+3}=x+y-\sqrt{x^2+3}-\sqrt{y^2+3}\)
\(\Leftrightarrow-y-x=x+y\Leftrightarrow2\left(x+y\right)=0\Leftrightarrow x+y=0\)
Vậy \(E=0.\)
