Question

Cho x,y thỏa mãn ( \(\sqrt{2+x^2}\) - x) (y + \(\sqrt{2+y^2}\)) = 2. CMR: x=y

Answer 1

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

\(\Leftrightarrow2\left(\sqrt{2+y^2}+y\right)=2\left(\sqrt{2+x^2}+x\right)\)

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

\(\Leftrightarrow\dfrac{\left(x-y\right)\left(x+y\right)}{\sqrt{2+x^2}+\sqrt{2+y^2}}+\left(x-y\right)=0\)

TH1:\(x-y=0\Leftrightarrow x=y\left(đpcm\right)\)

TH2: \(x+y+\sqrt{2+x^2}+\sqrt{2+y^2}=0\)

Ta có: \(x\ge-\sqrt{x^2}\); \(y\ge-\sqrt{y^2}\)

\(\Rightarrow x+y+\sqrt{2+x^2}+\sqrt{2+y^2}\ge\sqrt{2+x^2}-\sqrt{x^2}+\sqrt{2+y^2}-\sqrt{y^2}>0\)

Do vậy TH2 không có x,y tm

Vậy ta có đpcm