Đáp án:
Giải thích các bước giải:
$\dfrac{1}{x.(x+1)}+\dfrac{1}{(x+1).(x+2)}+\dfrac{1}{(x+2).(x+3)}-\dfrac{1}{x}=\dfrac{1}{2010}$
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$⇒\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}-\dfrac{1}{x}=\dfrac{1}{2010}$
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$⇒\dfrac{-1}{x+3}=\dfrac{1}{2010}$
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$⇒x+3=-2010$
$⇒x=-2013$