`(x + 4)/2018 + (x + 3)/2019 = (x + 2)/2020 + (x + 1)/2021`

`<=> (x + 4)/2018 + 1 + (x + 3)/2019 +1 = (x + 2)/2020 + 1 + (x + 1)/2021 +1`

`<=> (x + 2022)/2018 + (x + 2022)/2019 + (x + 2022)/2020 + (x + 2022)/2021`

`<=> (x + 2022)(1/2018 + 1/2019 - 1/2020 - 1/2021) = 0`

`<=> x + 2022 = 0` do `1/2018 + 1/2019 - 1/2020 - 1/2021 \ne 0`

`<=> x = -2022`

Vậy `x in {-2022}`