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如果a-1的绝对值+(b-2)的平方=0,求1/ab+1/(a+1)x(b+1)+1/(a+2)x(b+2)+…+1/(

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如果a-1的绝对值+(b-2)的平方=0,求1/ab+1/(a+1)x(b+1)+1/(a+2)x(b+2)+…+1/(a+2009)x(b+2009)的值
如果a-1的绝对值+(b-2)的平方=0,求1/ab+1/(a+1)x(b+1)+1/(a+2)x(b+2)+…+1/(
a=1,b=2
1/ab+1/(a+1)x(b+1)+1/(a+2)x(b+2)+…+1/(a+2009)x(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
=2010/2011