如果有理数a,b满足|ab-2|+(1-b)^2=0,试求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)…+1/(a+2004(b+2004)的值)

有理数a,b满足|ab-2|+(1-b)^2=0,
所以b=1,a=2
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)…+1/(a+2004(b+2004)
=1/2+1/2*3+1/3*4+1/4*5+.....+1/2005*2006
=1-1/2+1/2-1/3+1/3-1/4+...+1/2005-1/2006
=1-1/2006
=2005/2006

|ab-2|+(1-b)^2=0
则|ab-2|=(1-b)^2=0
则ab=2，b=1
则a=2，b=1
则1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)…+1/(a+2004(b+2004)
=1/(1*2)+1/(2*3)+...+1/(2005*2006)
=1/1-1/2+1/2-1/3+...+1/2005-1/2006
=1-1/2006
=2005/2006
请您给分吧