The set of real numbers $x$ for which
$$\dfrac{1}{x-2009}+\dfrac{1}{x-2010}+\dfrac{1}{x-2011}\ge 1$$is the union of intervals of the form $a<x\le b$. What is the sum of the lengths of these intervals?
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The set of real numbers $x$ for which
$$\dfrac{1}{x-2009}+\dfrac{1}{x-2010}+\dfrac{1}{x-2011}\ge 1$$is the union of intervals of the form $a<x\le b$. What is the sum of the lengths of these intervals?
Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
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user
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The set of real numbers $x$ for which
$$\dfrac{1}{x-2009}+\dfrac{1}{x-2010}+\dfrac{1}{x-2011}\ge 1$$is the union of intervals of the form $a<x\le b$. What is the sum of the lengths of these intervals?
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