Question

\(\left(x+\sqrt{x^2+2010}\right)\left(y+\sqrt{y^2+2010}\right)=2010\)

        tính x+y

     

   

 

Answer 1

 

\(\left(x+\sqrt{x^2+2010}\right)\left(y+\sqrt{y^2+2010}\right)=2010\)

\(\Leftrightarrow\left(x+\sqrt{x^2+2010}\right)\left(\sqrt{x^2+2010}-x\right)\left(y+\sqrt{y^2+2010}\right)=2010\left(\sqrt{x^2+2010}-x\right)\)

\(\Leftrightarrow y+\sqrt{y^2+2010}=\sqrt{x^2+2010}-x.\left(1\right)\)

Tương tự:\(x+\sqrt{x^2+2010}=\sqrt{y^2+2010}-y.\left(2\right)\)

Cộng vế với vế của \(\left(1\right)và\left(2\right)\Rightarrow x+y=-x-y\Leftrightarrow x+y=0.\)

KL:\(x+y=0.\)