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+2009)(b+2009)的值
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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+2009)(b+2009)的值](/uploads/image/z/3035784-48-4.jpg?t=%E5%B7%B2%E7%9F%A5%7Cab-2%7C%E4%B8%8E%7Cb-1%7C%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%2Fab%2B1%2F%28a%2B1%29%28b%2B1%29%2B1%2F%28a%2B2%29%28b%2B2%29%2B%E2%80%A6%2B1%2F%28a%2B2009%29%28b%2B2009%29%E7%9A%84%E5%80%BC)
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+2009)(b+2009)的值
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+2009)(b+2009)的值
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+2009)(b+2009)的值
|ab-2|+|b-1|=0,
所以|ab-2|=0,|b-1|=0,
所以ab-2=0,b-1=0,b=1,a=2,带入式子得
1/(1*2)+1/(2*3)+1/(3*4)+……+1/(2010*2011)
因为1/[n*(n+1)]=1/n-1/(n+1)
所以原式=(1-1/2)+(1/2-1/3)+……+(1/2010-1/2011)
=1-1/2011
=2010/2011.