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已知|ab-2|+(b+1)^2=0
求1/ab+1/(a-1)(b-1)+1/(a-2)(b-2)...+1/(a-2009)(b-2009)的值.
人气:489 ℃ 时间:2020-05-06 16:57:55
解答
已知|ab-2|+(b+1)^2=0
所以ab-2=0,b+1=0
所以b=-1,a=-2
所以
1/ab+1/(a-1)(b-1)+1/(a-2)(b-2)...+1/(a-2009)(b-2009)
=1/(-2)(-1)+1/(-3)(-2)+1/(-4)(-3)+...+1/(-2011)(-2010)
=1/1×2+1/2×3+1/3×4+...+1/2010×2011
=1-1/2+1/2-1/3+1/3-1/4+...+1/2010-1/2011
=1-1/2011
=2010/2011
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