\(\dfrac{x-3}{2014}+\dfrac{x-2}{2015}=\dfrac{x-2015}{2}+\dfrac{x-2014}{3}\)
⇔ \(\dfrac{x-3}{2014}-1+\dfrac{x-2}{2015}-1=\dfrac{x-2015}{2}-1+\dfrac{x-2014}{3}-1\)
⇔ \(\dfrac{x-2017}{2014}+\dfrac{x-2017}{2015}=\dfrac{x-2017}{2}+\dfrac{x-2017}{3}\)
⇔ \(\left(x-2017\right)\)\(\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)
Do : \(\dfrac{1}{2014}< \dfrac{1}{2015}< \dfrac{1}{2}< \dfrac{1}{3}\)
⇒ \(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2}-\dfrac{1}{3}< 0\)
⇔ x - 2017 = 0
⇔ x = 2017
KL....