求1/xy+1/(x+1)(y+1)+1/(x+2)(y+2)+…+1/(x+2009)(y+2009)+1/（x+2010）（y+2010）的值

如果 |x - 1| + (xy - 2)^2 = 0
那么|x - 1| = 0,xy - 2 = 0
从中解得 x = 1,y = 2
那么
1/xy + 1/(x+1)(y+1) + ··· + 1/(x+2010)(y+2010)
= 1 / (1·2) + 1 / (2·3) + ··· + 1/ (2011·2012)
= 1 - 1/2 + 1/2 - 1/3 +1/3 - ··· +1/2011 - 1/2012
= 1- 1/2012
= 2011 / 2012
做题思路：首先通过题目所给的式子可以求得x和y的值
后面利用1/[n(n+1)] = 1/n - 1/(n+1)这个裂项公式可以求出最后答案