Question

* Cho (x+\(\sqrt{x^2+3}\))\(\left(y+\sqrt{y^2+3}\right)=3\)

Tính x+y

Answer 1

Ta có: \(\left(x+\sqrt{x^2+3}\right)\left(y+\sqrt{y^2+3}\right)=3\)

mà \(\left(x+\sqrt{x^2+3}\right)\left(\sqrt{x^2+3}-x\right)=3\Rightarrow\sqrt{x^2+3}-x=y+\sqrt{y^2+3}\)

\(\Rightarrow x+y=\sqrt{x^2+3}-\sqrt{y^2+3}\left(1\right)\)

Mặt khác \(\left(y+\sqrt{y^2+3}\right)\left(\sqrt{y^2+3}-y\right)=3\)

\(\Rightarrow\sqrt{y^2+3}-y=\sqrt{x^2+3}+x\Rightarrow x+y=\sqrt{y^2+3}-\sqrt{x^2+3}\left(2\right)\)

Lấy \(\left(1\right)+\left(2\right)\Rightarrow2\left(x+y\right)=0\Rightarrow x+y=0\)