Đáp án:
`S=0`.
Giải thích các bước giải:
`(x+\sqrt{x^2+2021})(y+\sqrt{y^2+2021})=2021`
Ta có:
`(\sqrt{x^2+2021}-x)(x+\sqrt{x^2+2021})`
`=x^2+2021-x^2`
`=2021`
`=>(\sqrt{x^2+2021}-x)(x+\sqrt{x^2+2021})=(x+\sqrt{x^2+2021})(y+\sqrt{y^2+2021})=2021(1)`
`<=>\sqrt{x^2+2021}-x=y+\sqrt{y^2+2021}` do `x+\sqrt{x^2+2021}\ne0`
`<=>x+y=\sqrt{x^2+2021}-\sqrt{y^2+2021}(2)`
Hoàn toàn tương tự như (1) ta cũng có:
`(\sqrt{y^2+2021}-y)(y+\sqrt{y^2+2021})=(x+\sqrt{x^2+2021})(y+\sqrt{y^2+2021})=2021`
`<=>\sqrt{y^2+2021}-y=x+\sqrt{x^2+2021}` do `y+\sqrt{y^2+2021}\ne0`
`<=>x+y=\sqrt{y^2+2021}-\sqrt{x^2+2021}(3)`
Cộng từng vế (1)(2) ta có:
`2(x+y)=0<=>x+y=0`
`=>S=0`.