Ta có :
$\dfrac{x+2000}{20} +\dfrac{x+20}{2000} = \dfrac{x+1}{2019}+\dfrac{x+2}{2018} $
$⇔( \dfrac{x+2000}{20}+1) +(\dfrac{x+20}{2000} +1)= (\dfrac{x+1}{2019}+1)+(\dfrac{x+2}{2018}+1) $
$⇔ \dfrac{x+2020}{20} +\dfrac{x+2020}{2000} = \dfrac{x+2020}{2019}+\dfrac{x+2020}{2018} $
$⇔(x+2020).(\dfrac{1}{20} +\dfrac{1}{2000} - \dfrac{1}{2019}-\dfrac{1}{2018} ) = 0$
$⇔x+2020 = 0$
$⇔x=-2020$
Vậy pt có tập nghiệm $S = $ { $-2020 $ }
