Cho 2 số `x,y` thõa mãn đẳng thức `(x+\sqrt{x^2+2011})(y+\sqrt{y^2+2011})=2011` Tính `x+y` 19/07/2021 Bởi Parker Cho 2 số `x,y` thõa mãn đẳng thức `(x+\sqrt{x^2+2011})(y+\sqrt{y^2+2011})=2011` Tính `x+y`
Đáp án: $x+y=0$ Giải thích các bước giải: $(x+\sqrt{x^2+2011})(y+\sqrt{y^2+2011})=2011(*)$ $⇔(x+\sqrt{x^2+2011})(\sqrt{x^2+2011}-x)(y+\sqrt{y^2+2011})=2011(\sqrt{x^2+2011}-x)$ $⇔(x^2+2011-x^2)(y+\sqrt{y^2+2011})=2011(\sqrt{x^2+2011}-x)$ $⇔y+\sqrt{y^2+2011}=\sqrt{x^2+2011}-x(1)$ $(*)⇔(x+\sqrt{x^2+2011})(\sqrt{y^2+2011}-y)(y+\sqrt{y^2+2011})=2011(\sqrt{y^2+2011}-y)$ $⇔(y^2+2011-y^2)(x+\sqrt{x^2+2011})=2011(\sqrt{y^2+2011}-y)$ $⇔x+\sqrt{x^2+2011}=\sqrt{y^2+2011}-y(2)$ $(1);(2)⇒(x+\sqrt{x^2+2011})+(y+\sqrt{y^2+2011})=(\sqrt{y^2+2011}-y)+(\sqrt{x^2+2011}-x)$ $⇒x+y=-x-y⇒x+y=0$ Bình luận
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`. Bình luận
Đáp án: $x+y=0$
Giải thích các bước giải:
$(x+\sqrt{x^2+2011})(y+\sqrt{y^2+2011})=2011(*)$
$⇔(x+\sqrt{x^2+2011})(\sqrt{x^2+2011}-x)(y+\sqrt{y^2+2011})=2011(\sqrt{x^2+2011}-x)$
$⇔(x^2+2011-x^2)(y+\sqrt{y^2+2011})=2011(\sqrt{x^2+2011}-x)$
$⇔y+\sqrt{y^2+2011}=\sqrt{x^2+2011}-x(1)$
$(*)⇔(x+\sqrt{x^2+2011})(\sqrt{y^2+2011}-y)(y+\sqrt{y^2+2011})=2011(\sqrt{y^2+2011}-y)$
$⇔(y^2+2011-y^2)(x+\sqrt{x^2+2011})=2011(\sqrt{y^2+2011}-y)$
$⇔x+\sqrt{x^2+2011}=\sqrt{y^2+2011}-y(2)$
$(1);(2)⇒(x+\sqrt{x^2+2011})+(y+\sqrt{y^2+2011})=(\sqrt{y^2+2011}-y)+(\sqrt{x^2+2011}-x)$
$⇒x+y=-x-y⇒x+y=0$
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`.