Đáp án:`x+y=0`

 Còn cái chứng minh `x+sqrt{x^2+2011} ne 0` cũng khá dễ ta giả sử:

`x+sqrt{x^2+2011}=0`

`<=>-x=sqrt{x^2+2011}`

`đk:x<=0`

`<=>x^2=x^2+2011`

`<=>0=2011` vô lý.

`=>` Điều giả sử sai.

`=>x+sqrt{x^2+2011} ne 0`

Tương tự `y+sqrt{y^2+2011} ne 0`.

Giải thích các bước giải:

Ta có:

`(x+sqrt{x^2+2011})(sqrt{x^2+2011}-x)`

`=x^2+2011-x^2=2011`

Mà `(x+sqrt{x^2+2011})(y+sqrt{y^2+2011})=2011`

`=>(x+sqrt{x^2+2011})(sqrt{x^2+2011}-x)=(x+sqrt{x^2+2011}))(y+sqrt{y^2+2011})(1)`

`<=>(x+sqrt{x^2+2011})[sqrt{x^2+2011}-x-(y+sqrt{y^2+2011})]=0`

`<=>[sqrt{x^2+2011}-x-(y+sqrt{y^2+2011})]=0` do `x+sqrt{x^2+2011} ne 0`

`<=>sqrt{x^2+2011}-x=y+sqrt{y^2+2011}`

`<=>sqrt{x^2+2011}-sqrt{y^2+2011}=x+y(**)`

Tương tự như (1) ta có:

`(y+sqrt{y^2+2011})(sqrt{y^2+2011}-y)=(x+sqrt{x^2+2011})(y+sqrt{y^2+2011})`

`<=>(y+sqrt{y^2+2011})[sqrt{y^2+2011}-y-(x+sqrt{x^2+2011})]=0`

`<=>[sqrt{y^2+2011}-y-(x+sqrt{x^2+2011})]=0` do `y+sqrt{y^2+2011} ne 0`

`<=>sqrt{y^2+2011}-y=x+sqrt{x^2+2011}`

`<=>x+y=sqrt{y^2+2011}-sqrt{x^2+2011}(** **)`

Cộng từng vế `(**),(** **)` ta có:

`2(x+y)=sqrt{x^2+2011}-sqrt{y^2+2011}+sqrt{y^2+2011}-sqrt{x^2+2011}`

`<=>2(x+y)=0`

`<=>x+y=0`

Vậy với `(x+sqrt{x^2+2011})(y+sqrt{y^2+2011})=2011` thì `x+y=0`.

Đáp án: $x+y=0$

Giải thích các bước giải:

$(x+\sqrt{x^2+2011})(y+\sqrt{y^2+2011})=2011(*)$

$⇔(x+\sqrt{x^2+2011})(\sqrt{x^2+2011}-x)(y+\sqrt{y^2+2011})=2011(\sqrt{x^2+2011}-x)$

$⇔(x^2+2011-x^2)(y+\sqrt{y^2+2011})=2011(\sqrt{x^2+2011}-x)$

$⇔y+\sqrt{y^2+2011}=\sqrt{x^2+2011}-x(1)$

$(*)⇔(x+\sqrt{x^2+2011})(\sqrt{y^2+2011}-y)(y+\sqrt{y^2+2011})=2011(\sqrt{y^2+2011}-y)$

$⇔(y^2+2011-y^2)(x+\sqrt{x^2+2011})=2011(\sqrt{y^2+2011}-y)$

$⇔x+\sqrt{x^2+2011}=\sqrt{y^2+2011}-y(2)$

$(1);(2)⇒(x+\sqrt{x^2+2011})+(y+\sqrt{y^2+2011})=(\sqrt{y^2+2011}-y)+(\sqrt{x^2+2011}-x)$

$⇒x+y=-x-y⇒x+y=0$