Question

Cho: \(\left(x+\sqrt{x^2+5}\right)\left(y+\sqrt{y^2+5}\right)=5\)

Tính \(H=x+y\)

Answer 1

\(\left(x-\sqrt{x^2+5}\right)\left(x+\sqrt{x^2+5}\right)\left(y+\sqrt{y^2+5}\right)=5\left(x-\sqrt{x^2+5}\right)\)

<=> \(-5\left(y+\sqrt{y^2+5}\right)=5\left(x-\sqrt{x^2+5}\right)\Leftrightarrow-y-\sqrt{y^2+5}=x-\sqrt{x^2+5}\left(1\right)\)

Tương tự \(-x-\sqrt{x^2+5}=y-\sqrt{y^2+5}\left(2\right)\)

Từ (1) và (2) => \(-x-y-\sqrt{x^2+5}-\sqrt{y^2+5}=x+y-\sqrt{x^2+5}-\sqrt{y^2+5}\)

<=> \(-x-y=x+y\Leftrightarrow2\left(x+y\right)=0\Rightarrow x+y=0\Leftrightarrow H=0\)