\(\dfrac{x-3}{2011}+\dfrac{x-2}{2012}=\dfrac{x-2012}{2}+\dfrac{x-2011}{3}\)

\(\Leftrightarrow\left(\dfrac{x-3}{2011}-1\right)+\left(\dfrac{x-2}{2012}-1\right)=\left(\dfrac{x-2012}{2}-1\right)+\left(\dfrac{x-2011}{3}-1\right)\)

\(\Leftrightarrow\left(\dfrac{x-3-2011}{2011}\right)+\left(\dfrac{x-2-2012}{2012}\right)=\left(\dfrac{x-2012-2}{2}\right)+\left(\dfrac{x-2011-3}{3}\right)\)

\(\Leftrightarrow\left(\dfrac{x-2014}{2011}\right)+\left(\dfrac{x-2014}{2012}\right)=\left(\dfrac{x-2014}{2}\right)+\left(\dfrac{x-2014}{3}\right)\)

\(\Leftrightarrow\left(\dfrac{x-2014}{2011}\right)+\left(\dfrac{x-2014}{2012}\right)-\left(\dfrac{x-2014}{2}\right)-\left(\dfrac{x-2014}{3}\right)=0\)

\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)

\(\Leftrightarrow\left(x-2014\right)=0\) ( vì \(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2}-\dfrac{1}{3}e0\) )

\(\Leftrightarrow x=2014\)

Vậy phương trình có nghiệm \(S=\left\{2014\right\}\)