$\dfrac{3-x}{2019}-\dfrac{2-x}{2020}+\dfrac{1-x}{2021}=-1$
$⇔\dfrac{3-x}{2019}+1-\dfrac{2-x}{2020}-1+\dfrac{1-x}{2021}+1=0$
$⇔\dfrac{2022-x}{2019}-\dfrac{2022-x}{2020}+\dfrac{2022-x}{2021}=0$
$⇔(2022-x)(\dfrac{1}{2019}-\dfrac{1}{2020}+\dfrac{1}{2021})=0$
$⇔2022-x=0$
$⇔x=2022$