Question

[Quách Trần Gia Lạc]

Cho \(\left(x+\sqrt{x^2+2007}\right)\left(y+\sqrt{y^2+2007}\right)=2007\). Tính S = x + y.

Answer 1

[Akai Haruma]

Lời giải:
Ta có:

\((x+\sqrt{x^2+2007})(y+\sqrt{y^2+2007})=2007\)

Nhân \(x-\sqrt{x^2+2007}\) vào 2 vế:

\(\Rightarrow (x-\sqrt{x^2+2007})(x+\sqrt{x^2+2007})(y+\sqrt{y^2+2007})=2007(x-\sqrt{x^2+2007})\)

\(\Leftrightarrow [x^2-(x^2+2007)](y+\sqrt{y^2+2007})=2007(x-\sqrt{x^2+2007})\)

\(\Leftrightarrow -2007(y+\sqrt{y^2+2007})=2007(x-\sqrt{x^2+2007})\)

\(\Leftrightarrow -(y+\sqrt{y^2+2007})=x-\sqrt{x^2+2007}\)

\(\Leftrightarrow x+y+\sqrt{y^2+2007}-\sqrt{x^2+2007}=0(1)\)

Hoàn toàn tương tự, nhân \(y-\sqrt{y^2+2007}\) vào 2 vế:

\(x+y+\sqrt{x^2+2007}-\sqrt{y^2+2007}=0(2)\)

Từ (1);(2) suy ra: \(2(x+y)=0+0=0\Rightarrow S=x+y=0\)