\(\dfrac{x-3}{2011}+\dfrac{x-2}{2012}=\dfrac{x-2012}{2}+\dfrac{x-2011}{3}\)
\(\Leftrightarrow\left(\dfrac{x-3}{2011}-1\right)+\left(\dfrac{x-2}{2012}-1\right)=\left(\dfrac{x-2012}{2}-1\right)+\left(\dfrac{x-2011}{3}-1\right)\)
\(\Leftrightarrow\left(\dfrac{x-3-2011}{2011}\right)+\left(\dfrac{x-2-2012}{2012}\right)=\left(\dfrac{x-2012-2}{2}\right)+\left(\dfrac{x-2011-3}{3}\right)\)
\(\Leftrightarrow\left(\dfrac{x-2014}{2011}\right)+\left(\dfrac{x-2014}{2012}\right)=\left(\dfrac{x-2014}{2}\right)+\left(\dfrac{x-2014}{3}\right)\)
\(\Leftrightarrow\left(\dfrac{x-2014}{2011}\right)+\left(\dfrac{x-2014}{2012}\right)-\left(\dfrac{x-2014}{2}\right)-\left(\dfrac{x-2014}{3}\right)=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)
\(\Leftrightarrow\left(x-2014\right)=0\) ( vì \(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2}-\dfrac{1}{3}\ne0\) )
\(\Leftrightarrow x=2014\)
Vậy phương trình có nghiệm \(S=\left\{2014\right\}\)
\(\dfrac{x-3}{2011}+\dfrac{x-2}{2012}=\dfrac{x-2012}{2}+\dfrac{x-2011}{3}\)
\(\Leftrightarrow\dfrac{x-3}{2011}-1+\dfrac{x-2}{2012}-1=\dfrac{x-2012}{2}-1+\dfrac{x-2011}{3}-1\)
\(\Leftrightarrow\dfrac{x-3-2011}{2011}+\dfrac{x-2-2012}{2012}=\dfrac{x-2012-2}{2}+\dfrac{x-2011-3}{3}\)
\(\Leftrightarrow\dfrac{x-2014}{2011}+\dfrac{x-2014}{2012}-\dfrac{x-2014}{2}-\dfrac{x-2014}{3}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)
\(\Leftrightarrow x-2014=0\) ( Vì: \(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2}-\dfrac{1}{3}\ne0\))
Vậy x = 2014