Question

Cho \(\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1\)

Tính A=\(x^{2017}+y^{2017}\)

Answer 1

Ta có:

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

<=> \(y+\sqrt{y^2+1}=\sqrt{x^2+1}-x\)

<=> \(x+y=\sqrt{x^2+1}-\sqrt{y^2-1}\) (1)

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

<=> \(x+\sqrt{x^2+1}=\sqrt{y^2+1}-y\)

<=> \(x+y=\sqrt{y^2+1}-\sqrt{x^2+1}\) (2)

Cộng (1) và (2) vế theo vế ta được:

\(2\left(x+y\right)=0\) <=> \(x+y=0\) <=> \(x=-y\)

Thay \(x=-y\) vào \(x^{2017}+y^{2017}\) ta có:

A = \(\left(-y\right)^{2017}+y^{2017}=0\)