Question

Tính x + y ?

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

Answer 1

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

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

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

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

Trừ vế theo vế (1) và (2)

\(\Rightarrow2x=-2y\Leftrightarrow2x+2y=0\Leftrightarrow2\left(x+y\right)=0\Leftrightarrow x+y=0\)