已知a,b,c,d满足(a-2001)(b+2001)+(c+2001)(d-2002)=1 (a-2002)(b+20

(a-2001)(b+2001)+(c+2001)(d-2001)=1 ----原题目是d-2002,应该是d-2001吧?否则无法做
(a-2002)(b+2002)+(c+2002)(d-2002)=10
两者相减得到
a -b -2002 - 2001 + d-c - (2002 +2001) =9
a-b +d-c = 9 +2 (2002 +2001) -----1
(a-2011)(b+2011) + (c+2011)(d-2011) - [(a-2001)(b+2001)+(c+2001)(d-2002)]=
10a - 10b -2011^2 + 2001^2 +10d-10c-2011^2 + 2001^2 =
10(a-b+d-c) -2(2011^2 -2001^2)
带进1）就可以得到答案了