`(x+1)/2019+(x+2)/2018=(x+3)/2017+(x+4)/2016`
`<=>((x+1)/2019+1)+((x+2)/2018+1)=((x+3)/2017+1)+((x+4)/2016+1)`
`<=>(x+2020)/2019+(x+2020)/2018=(x+2020)/2017+(x+2020)/2016`
`<=>(x+2020)/2019+(x+2020)/2018-(x+2020)/2017-(x+2020)/2016=0`
`<=>(x+2020)(1/2019+1/2018-1/2017-1/2016)=0`
Nhận thấy:
`1/2019<1/2017=>1/2020-1/2017<0`
`1/2018<1/2016=>1/2018-1/2016<0`
`=>1/2019+1/2018-1/2017-1/2016<0`
`=>x+2020=0`
`<=>x=-2020`
Vậy `S={-2020}`
