Question

x y为实数 且x²-2x+√xy-2=-1 求1/xy+1/﹙x+1﹚﹙y+1﹚+.....+1/﹙x+2006﹚﹙y+2006﹚的值

麻烦了

Answer 1

(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+2006﹚﹙y+2006﹚
=1/1*2+1/2*3+1/3*4+……+1/2007*2008
=1-1/2+1/2-1/3+1/3-……-1/2007+1/2007-1/2008
=1-1/2008
=2007/2008

追问

(x-1)²+√(xy-2)=0
怎么来的

追答

x²-2x+√xy-2=-1
x²-2x+1+√xy-2=0
(x-1)²+√(xy-2)=0