\(\dfrac{2-x}{2017}-1=\dfrac{1-x}{2018}-\dfrac{x}{2019}\\↔\dfrac{2-x}{2017}+1-2=\dfrac{1-x}{2018}-\dfrac{x}{2019}\\↔\dfrac{2019-x}{2017}=\dfrac{1-x}{2018}+1-\dfrac{x}{2019}+1\\↔\dfrac{2019-x}{2017}=\dfrac{2019-x}{2018}+\dfrac{2019-x}{2019}\\↔\dfrac{2019-x}{2017}-\dfrac{2019-x}{2018}-\dfrac{2019-x}{2019}=0\\↔(2019-x)\bigg(\dfrac{1}{2017}-\dfrac{1}{2018}-\dfrac{1}{2019}\bigg)=0\\↔2019-x=0\bigg(vi\,\, \dfrac{1}{2017}-\dfrac{1}{2018}-\dfrac{1}{2019}\ne 0\bigg)\\↔x=2019\)
Vậy pt có tập nghiệm \(S=\{2019\}\)
