Question

若a-1的绝对值+ab-2的绝对值=0，求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.......1/(a+2002)(b+2002)

Answer 1

因为|a-1|+|ab-2|的2次方=0，所以两个都等于0
所以a=1，b=2
根据分解公式：
1/ab+1/(a+1)(b+1)+...+1/(a+2002)(b+2002)
=1/(1*2)+1/(2*3)+...+1/(2003*2004)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+....+(1/2002-1/2003)+(1/2003-1/2004)
=1-1/2004
=2003/2004

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Answer 2

｜ab-2｜+｜b-1｜=0
绝对值相加为0则都等于0
所以ab-2=0，b-1=0
b=1,ab=2,所以a=2
所以
1/ab+1/(a+1)(b+1)+…+1/(a+2003)(b+2003)
=1/1*2+1/2*3+……+1/2004*2005
=1-1/2+1/2-1/3+……+1/2004-1/2005
=1-1/2005
=2004/2005