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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+20
人气:448 ℃ 时间:2020-03-26 17:29:38
解答
∵|ab-2|与|b-1|互为相反数,∴|ab-2|+|b-1|=0,∴ab-2与b-1同时等于0,∴ab=2,b=1,∴a=2,b=1.
∴1/ab+1/(a+B)(b+1)+1/(a+2)(b+2)+...+1/(a+2009)(b+2009)
=1/2 + 1/3乘以2+ 1/4乘以3 + .+1/2011乘以2010
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+.+(1/2010-1/2011)(拆一项成两项的差得来的)
=1 - 1/2011
=2010/2011
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