|a+1|与|b-2|互为相反数,求代数式:1/(a+2)b+1/(a+3)(b+1)+...+1/(a+2010)(b+2008)的值
人气:178 ℃ 时间:2020-01-27 07:08:05
解答
|a+1|与|b-2|互为相反数
则|a+1|+|b-2|=0
则a+1=0,b-2=0
则a=-1,b=2
1/(a+2)b+1/(a+3)(b+1)+...+1/(a+2010)(b+2008)
=1/(1*2)+1/(2*3)+…+1/(2008*2009)
=1-1/2+1/2-1/3+…+1/2008-1/2009
=1-1/2009
=2008/2009
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