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
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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](/uploads/image/z/3876763-67-3.jpg?t=%E5%B7%B2%E7%9F%A5%EF%B8%B4ab-2%EF%B8%B4%E4%B8%8E%EF%B8%B4b-1%EF%B8%B4%E4%BA%92%E4%B8%BA%E7%9B%B8%E5%8F%8D%E6%95%B0%2C%E8%AF%95%E6%B1%82%E4%BB%A3%E6%95%B0%E5%BC%8F1%E2%88%95ab%2B1%2F%28a%2B1%29%28b%2B1%29%2B1%2F%28a%2B2%29%28b-2%29%2B%E2%80%A6%2B1%2F%EF%BC%88a%2B20)
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
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
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
∵|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
互为相反数则相加为0
|ab-2|=|b-1|=0
绝对值大于等于0
相加等于0,若有一个大于0,则另一个小于0,不成立。
所以两个都等于0
所以ab=2,b=1
所以a=2/b=2
所以
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+2008)(b+2008
=1/1*2+1/2*3+1/3*...
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互为相反数则相加为0
|ab-2|=|b-1|=0
绝对值大于等于0
相加等于0,若有一个大于0,则另一个小于0,不成立。
所以两个都等于0
所以ab=2,b=1
所以a=2/b=2
所以
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+2008)(b+2008
=1/1*2+1/2*3+1/3*4+……+1/2009*2010
=1-1/2+1/2-1/3+1/3-1/4+……+1/2009-1/2010
中间正负抵消
=1-1/2010
=2009/2010
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