已知|a-1|与|ab-2|互为相反数,求下列代数式的值:1/ab+1/(a+1)(b+1)+...+1/(a+2002)(b+2002)
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人气:290 ℃ 时间:2020-02-05 06:29:06
解答
|a-1|与|ab-2|都是非负数
又互为相反数,实际上它们都为0
所以a=1,b=2
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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