Question

cho \(\left(x+\sqrt{x^2+2019}\right)\left(y+\sqrt{y^2+2019}\right)=2019\). CM: \(x^{2019}+y^{2019}=0\)

Answer 1

Từ gt suy ra: \(x+\sqrt{x^2+2019}=\dfrac{2019}{y+\sqrt{y^2+2019}}=\sqrt{y^2+2019}-y\).

Tương tự: \(y+\sqrt{y^2+2019}=\sqrt{x^2+2019}-x\).

Do đó dễ dàng suy ra được: \(x+y=0\).

\(\Rightarrow x=-y\Rightarrow x^{2019}+y^{2019}=x^{2019}+\left(-x\right)^{2019}=0\left(đpcm\right)\).