题目：x、y为实数,满足(x-1)³+1999(x-1)=-1 ,(y-1)³+1999(y-1)=1,求x+y.
将两已知等式相加,得：
(x-1)³+1999(x-1)+(y-1)³+1999(y-1)=0
[(x-1)³+(y-1)³]+[1999(x-1)+1999(y-1)]=0
[(x-1)+(y-1)]*[(x-1)²+(y-1)²-(x-1)(y-1)]+1999*[(x-1)+(y-1)]=0
(x+y-2)*[(x-1)²+(y-1)²-(x-1)(y-1)]+1999*(x+y-2)=0
(x+y-2)*[(x-1)²+(y-1)²-(x-1)(y-1)+1999]=0
可得：x+y-2=0,得：x+y=2；
对于中括号内的数是不等于0的,理由如下：
(x-1)²+(y-1)²-(x-1)(y-1)+1999
=x²-2x+1+y²-2y+1-(xy-x-y+1)+1999
=x²+y²-xy-x-y+2000
=(1/2x²+1/2y²-xy)+(1/2x²-x+1/2)+(1/2y²-y+1/2)+1999
=1/2(x²+y²-2xy)+1/2(x²-2x+1)+1/2(y²-2y+1)+1999
=1/2(x-y)²+1/2(x-1)²+1/2(y-1)²+1999>0
综上,只能是：x+y=2.