Question

已知（X-1）²+（y-2）²=0

求1/XY+1/(x+1)(Y+1)+1/(X+2)(Y+2)+······+1/（x+1989）（y+1989）的值

Answer 1

由条件可知x=1，y=2，则原式=1/(1*2)+1/(2*3)+……+1/(1990*1991)。
采用裂项法，1/(1*2)=1/1-1/2，1/(2*3)=1/2-1/3，因此原式=1/1-1/2+1/2-1/3+1/3-1/4+……+1/1990-1/1991=1-1/1991=1990/1991。

Answer 2

（X-1）²+（y-2）²=0
∴x-1=0
y-2=0
∴x=1
y=2
1/XY+1/(x+1)(Y+1)+1/(X+2)(Y+2)+······+1/（x+1989）（y+1989）
=1/1×2+1/2×3+1/3×4+……+1/1990×1991
=1-1/2+1/2-1/3+1/3-1/4+……+1/1990-1/1991
=1-1/1991
=1990/1991

Answer 3

∵(X-1）²+（y-2）²=0
∴x-1=0，y-2=0
∴x=1，y=2

1/XY+1/(x+1)(Y+1)+1/(X+2)(Y+2)+······+1/（x+1989）（y+1989）
=1/1*2+1/2*3+1/3*4+……+1/1990*1991
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+……+(1/1990-1/1991)
=1-1/2+1/2-1/3+1/3-1/4+……+1/1990-1/1991
=1-1/1991
=1990/1991

Answer 4

1990/1991