Ta có \(\left(x+\sqrt{x^2+3}\right)\left(y+\sqrt{y^2+3}\right)=3\)
Mà \(-\left(x+\sqrt{x^2+3}\right)\left(x-\sqrt{x^2+3}\right)=-\left(x^2-x^2-3\right)=3\)
Suy ra \(\sqrt{x^2+3}-x=y+\sqrt{y^2+3}\)(1)
Chứng minh tương tự \(x+\sqrt{x^2+3}=\sqrt{y^2+3}-y\)(2)
Trừ vế theo vế của (1) cho (2)\(\Rightarrow2x=-2y\Leftrightarrow2x+2y=0\Leftrightarrow2\left(x+y\right)=0\Leftrightarrow x+y=0\)
CÁCH KHÁC:
\(\Leftrightarrow\frac{x^2-x^2-3}{x-\sqrt{x^2+3}}\left(y+\sqrt{y^2+3}\right)=3\)
\(\Leftrightarrow\frac{-1}{x-\sqrt{x^2+3}}\left(y+\sqrt{y^2+3}\right)=1\)
\(\Rightarrow-y-\sqrt{y^2+3}=x-\sqrt{x^2+3}\)
\(\Leftrightarrow x+y=\sqrt{x^2+3}-\sqrt{y^2+3}\left(1\right)\)
*pt\(\Leftrightarrow\left(x+\sqrt{x^2+3}\right).\frac{-3}{y-\sqrt{y^2+3}}=3\)
\(\Leftrightarrow\frac{-x-\sqrt{x^2+3}}{y-\sqrt{y^2+3}}=1\)
\(\Rightarrow y-\sqrt{y^2+3}=-x-\sqrt{x^2+3}\)
\(\Leftrightarrow x+y=\sqrt{y^2+3}-\sqrt{x^2+3}\left(2\right)\)
Từ (1)(2),có: \(\left\{{}\begin{matrix}x+y=\sqrt{x^2+3}-\sqrt{y^2+3}\\x+y=-\sqrt{x^2+3}+\sqrt{y^2+3}\end{matrix}\right.\)
\(\Rightarrow2\left(x+y\right)=0\)
\(\Leftrightarrow x+y=0\)
