已知|ab-2|与|b-1|互为相反数,试求代数式1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/11/08 06:30:44
已知|ab-2|与|b-1|互为相反数,试求代数式1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2012)
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2012)(b+2012)
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2012)(b+2012)
|ab-2|+|b-1|=0
b-1=0 b=1
ab=2 a=2/b=2
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2012)(b+2012)
=1/1x2+1/2x3+1/3x4+...+1/2013x2014
=1-1/2+1/2-1/3+1/3-1/4+...+1/2013-1/2014
=1-1/2014
=2013/2014
b-1=0 b=1
ab=2 a=2/b=2
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2012)(b+2012)
=1/1x2+1/2x3+1/3x4+...+1/2013x2014
=1-1/2+1/2-1/3+1/3-1/4+...+1/2013-1/2014
=1-1/2014
=2013/2014
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