设x,y是实数,且满足(x-1)^3+2008(x-1)=-1,(y-1)^3+2008(y-1)=1,则x+y=
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x+y=2
(x-1)^3+2008(x-1)=-1
(y-1)^3+2008(y-1)=1
(x-1)^3+2008(x-1)+(y-1)^3+2008(y-1)=0
(x-1)^3+(y-1)^3+2008(x-1+y-1)=0
((x-1)+(y-1))((x-1)^2-(x-1)(y-1)+(y-1)^2)+2008(x+y-2)=0
(x+y-2)((x-1)^2-(x-1)(y-1)+(y-1)^2+2008)=0
(x+y-2)(((x-1)-(y-1)/2)^2+3(y-1)^2/4+2008)=0
(x+y-2)((x+y-2)^2+3(y-1)^2/4+2008)=0
(x+y-2)^2+3(y-1)^2/4+2008>0
所以 x+y-2=0
x+y=2
(x-1)^3+2008(x-1)=-1
(y-1)^3+2008(y-1)=1
(x-1)^3+2008(x-1)+(y-1)^3+2008(y-1)=0
(x-1)^3+(y-1)^3+2008(x-1+y-1)=0
((x-1)+(y-1))((x-1)^2-(x-1)(y-1)+(y-1)^2)+2008(x+y-2)=0
(x+y-2)((x-1)^2-(x-1)(y-1)+(y-1)^2+2008)=0
(x+y-2)(((x-1)-(y-1)/2)^2+3(y-1)^2/4+2008)=0
(x+y-2)((x+y-2)^2+3(y-1)^2/4+2008)=0
(x+y-2)^2+3(y-1)^2/4+2008>0
所以 x+y-2=0
x+y=2
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