若|ab-2|+|b-1|=0,试求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2003)(b+2003)若|ab-2|+|b-1|=0,试求 1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2003)(b+2003)

若|ab-2|+|b-1|=0,试求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2003)(b+2003)若|ab-2|+|b-1|=0,试求 1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2003)(b+2003)
若|ab-2|+|b-1|=0,试求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2003)(b+2003)
若|ab-2|+|b-1|=0,试求 1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2003)(b+2003)

若|ab-2|+|b-1|=0,试求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2003)(b+2003)若|ab-2|+|b-1|=0,试求 1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2003)(b+2003)
易知ab=2 b=1 所以a=2 所以原式子就变成了1/2+1/2*3+1/3*4+.+1/2004*2005=1-1/2005=2004/2005 这个解法你应该会吧 有这么一个式子就是 1/（n+1)n=1/n-1/(n+1)