Ta có:\(\left(x+\sqrt{x^2+2003}\right)\left(y+\sqrt{y^2+2003}\right)=\dfrac{\left(x+\sqrt{x^2+2003}\right)\left(x-\sqrt{x^2+2003}\right)\left(y+\sqrt{y^2+2003}\right)\left(y-\sqrt{y^2+2003}\right)}{\left(x-\sqrt{x^2+2003}\right)\left(y-\sqrt{y^2+2003}\right)}=\dfrac{\left(x^2-x^2-2003\right)\left(y^2-y^2-2003\right)}{\left(x-\sqrt{x^2+2003}\right)\left(y-\sqrt{y^2+2003}\right)}=\dfrac{2003^2}{\left(x-\sqrt{x^2+2003}\right)\left(y-\sqrt{y^2+2003}\right)}=2003\)
=>\(\left(x-\sqrt{x^2+2003}\right)\left(y-\sqrt{y^2+2003}\right)=2003\)
=>\(\left(x-\sqrt{x^2+2003}\right)\left(y-\sqrt{y^2+2003}\right)=\left(x+\sqrt{x^2+2003}\right)\left(y+\sqrt{y^2+2003}\right)\)
nhân phá và thu gọn ta được
\(x\sqrt{y^2+2003}=-y\sqrt{x^2+2003}\)(1)
Bình phương
=>x2y2+2003x2=x2y2+2003y2
<=>x2=y2
<=>x=y hoặc x=-y
Thay vào (1) thì
x=y <=>x=y=0
x=-y (luôn đúng)
=>x+y=0
