若|ab-2|+|b-1|=0,试求ab/1+(a+1)(b+1)/1+(a+2)(b+2)+…+(a+2 012)(b

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若|ab-2|+|b-1|=0,试求ab/1+(a+1)(b+1)/1+(a+2)(b+2)+…+(a+2 012)(b+2 012)的值.

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已知(a-1)的平方+ab-2的绝对值=0 求ab/1+(a+1)(b+1)/1+(a+2)(b+2)/1一直这样加到
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因为(a-1)^2>=0
|ab-2|>=0
欲使(a-1)^2+|ab-2|=0
所以必有
(a-1)^2=0
|ab-2|=0
可得:a=1,b=2
所以1/ab+1/[(a+1)(b+1)]+1/[(a+2)(b+2)]+...+1/[(a+2009)(b+2009)]
=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
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|a-b-1|与|b-1|互为相反数
∴ |a-b-1|+|b-1|=0
∵ 绝对值非负.
∴ a-b-1=0,b-1=0
∴ b=1,a=2
∴ ab/1+(a+1)(b+1)/1+(a+2)(b+2)/1+…+(a+2008)(b+2008)
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因为|A - 1| >= 0,(AB-2)² >= 0
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所以|A - 1| = 0,(AB-2)² = 0
A = 1,AB = 2
A = 1,B = 2
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