若x,y为实数,且x的平方-2x+根号下xy-2=-1你能求出xy分之1+（x+1）（y+1）分之一+（x+2）（y+2）分之一+…+（x+2010）（y+2010）的值吗?

题意得
x²-2x+1+√(xy-2)=0
(x-1)²+√(xy-2)=0
∴x-1=0 xy-2=0
∴x=1 y=2
∴原式=1/(1×2)+1/(2×3)+1/(3×4)+....+1/(2011×2012)
=1-1/2+1/2-1/3+1/2-1/4+...+1/2011-1/2012
=1-1/2012
=2111/2012

x²-2x+√xy-2=-1
x²-2x+1+√xy-2=0
(x-1)²+√(xy-2)=0∴x-1=0
xy-2=0
∴x=1
y=2
1/xy+1/﹙x+1﹚﹙y+1﹚+.+1/﹙x+2010﹚﹙y+2010﹚
=1/1*2+1/2*3+1/3*4+……+1/2011*2012
=1-1/2+1/2-1/3+1/3-……-1/2011+1/2012-1/2012
=1-1/2012
=2011/2012