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24 tháng 3 2022 lúc 7:52

\(\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}=\dfrac{x-4}{2010}+\dfrac{x-5}{2009}+\dfrac{x-6}{2008}\)

\(\Leftrightarrow\left(\dfrac{x-1}{2013}-1\right)+\left(\dfrac{x-2}{2012}-1\right)+\left(\dfrac{x-3}{2011}-1\right)-\left(\dfrac{x-4}{2010}-1\right)-\left(\dfrac{x-5}{2009}-1\right)-\left(\dfrac{x-6}{2008}-1\right)=0\)

\(\Leftrightarrow\left(\dfrac{x-2014}{2013}\right)+\left(\dfrac{x-2014}{2012}\right)+\left(\dfrac{x-2014}{2011}\right)-\left(\dfrac{x-2014}{2010}\right)-\left(\dfrac{x-2014}{2009}\right)-\left(\dfrac{x-6}{2008}\right)=0\)

\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2008}\right)=0\)

\(\Leftrightarrow x-2014=0\)

\(\Leftrightarrow x=2014\)

Vậy \(S=\left\{2014\right\}\)

 

 

 

 

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