\(\dfrac{x-3}{2011}+\dfrac{x-2}{2011}=\dfrac{x-2011}{2}+\dfrac{x-2011}{3}\)
\(\Rightarrow (\dfrac{x-3}{2011}-1)+(\dfrac{x-2}{2011}-1)=(\dfrac{x-2011}{2}-1)+(\dfrac{x-2011}{3}-1)\)
\(\Rightarrow \dfrac{x-2014}{2011}+\dfrac{x-2014}{2012}=\dfrac{x-2014}{2}+\dfrac{x-20114}{3}\)
\(\Rightarrow \dfrac{x-2014}{2011}+\dfrac{x-2014}{2012}-\dfrac{x-2014}{2}-\dfrac{x-2014}{3}=0\)
\(\Rightarrow (x-2014).(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2}-\dfrac{1}{3})=0\)
\(Vì \) \(\dfrac{1}{2011}+\dfrac{1}{3012}-\dfrac{1}{2}-\dfrac{1}{3}\) \(\ne\)\(0\)
\(\Rightarrow x-2014=0\)
\(\Rightarrow x= 2014\)
\(Vậy \) \(x=2014\)
x-3/2011 + x-2/2012 = x-2012/3 + x-2011/2
<=> x-3/2011 +1 + x-2/2012 +1 = x-2012/3 + 1 + x-2011/2 +1
<=> x-2008/2011 + x-2008/2012 = x-2008/3 + x-2008/2
<=> x-2008/2011 + x-2008/2012 - x-2008/3 - x-2008/ 2 =0
<=> (x-2008) (1/2011 + 1/2012 - 1/3 -1/2) =0
<=> x- 2008 =0 (do 1/2011 +1/2012 - 1/3 - 1/2 ≠ 0)
<=> x= 2008
Vạy phương trình đã cho có nghiệm x=2008
mình cũng không chắc nữa ^^
