\(x-\dfrac{1}{2017}+x-\dfrac{2}{2016}=x-\dfrac{3}{2015}+x-\dfrac{4}{2014}\)
\(\Rightarrow x+x-x-x=-\dfrac{4}{2014}-\dfrac{3}{2015}+\dfrac{2}{2016}+\dfrac{1}{2017}\)
\(\Rightarrow0=-\dfrac{4}{2014}-\dfrac{3}{2015}+\dfrac{2}{2016}+\dfrac{1}{2017}\) (vô lí)
\(\dfrac{x-1}{2017}+\dfrac{x-2}{2016}=\dfrac{x-3}{2015}+\dfrac{x-4}{2014}\Leftrightarrow\left(\dfrac{x-1}{2017}-1\right)+\left(\dfrac{x-2}{2016}-1\right)=\left(\dfrac{x-3}{2015}-1\right)+\left(\dfrac{x-4}{2014}-1\right)\Leftrightarrow\left(x-2018\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\Leftrightarrow x=2018\)
Ta có: \(\dfrac{x-1}{2017}+\dfrac{x-2}{2016}=\dfrac{x-3}{2015}+\dfrac{x-4}{2014}\)
\(\Leftrightarrow\dfrac{x-2018}{2017}+\dfrac{x-2018}{2016}-\dfrac{x-2018}{2015}-\dfrac{x-2018}{2014}=0\)
\(\Leftrightarrow x-2018=0\)
hay x=2018
