\(\left(x+\sqrt{x^2+2020}\right)\left(y+\sqrt{y^2+2020}\right)=2020\)
\(\Leftrightarrow\hept{\begin{cases}\frac{2020}{x+\sqrt{x^2+2020}}=y+\sqrt{y^2+2020}\\\frac{2020}{y+\sqrt{y^2+2020}}=x+\sqrt{x^2+2020}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}-x+\sqrt{x^2+2020}=y+\sqrt{y^2+2020}\\-y+\sqrt{y^2+2020}=x+\sqrt{x^2+2020}\end{cases}}\)
\(\Leftrightarrow-2x-2y=0\)(cộng 2 vế )
\(\Leftrightarrow x+y=0\)
Mềnh còn cách khác:)
\(\left(x+\sqrt{x^2+2020}\right)\left(y+\sqrt{y^2+2020}\right)=2020\)
Ta có:\(\left(\sqrt{x^2+2020}+x\right)\left(\sqrt{x^2+2020}-x\right)=x^2+2020-x^2=2020\)
Lại có:\(\left(\sqrt{x^2+2020}+x\right)\left(\sqrt{y^2+2020}+y\right)=2020\)
\(\Rightarrow\sqrt{x^2+2020}-x=\sqrt{y^2+2020}+y\)
\(\Leftrightarrow x+y=\sqrt{x^2+2020}-\sqrt{y^2+2020}\)(1)
\(\left(\sqrt{y^2+2020}+y\right)\left(\sqrt{y^2+2020}-y\right)=y^2+2020-y^2=2020\)
\(\Rightarrow\sqrt{y^2+2020}-y=\sqrt{x^2+2020}+x\)
\(\Leftrightarrow x+y=\sqrt{y^2+2020}-\sqrt{x^2+2020}\)(2)
Cộng vế với vế của (1) và (2) ta có:\(x+y+x+y=\sqrt{x^2+2020}-\sqrt{y^2+2020}+\sqrt{y^2+2020}-\sqrt{x^2+2020}\)
\(\Leftrightarrow2x+2y=0\Leftrightarrow2\left(x+y\right)=0\Leftrightarrow x+y=0\)
