Đáp án+Giải thích các bước giải:
`(x-1)/2021+(x-2)/2020+ (x-3)/2019=(x-4)/2018 +(x-5)/2017+(x-6)/2016`
`↔((x-1)/2021-1)+((x-2)/2020-1)+ ((x-3)/2019-1)=((x-4)/2018-1) +((x-5)/2017-1)+((x-6)/2016-1)`
`↔(x-2022)/2021+(x-2022)/2020+(x-2022)/2019=(x-2022)/2018 +(x-2022)/2017+(x-2022)/2016`
`↔(x-2022)(1/2021+1/2020+1/2019-1/2018-1/2017-1/2016)=0`
`↔x=2022` (vì `1/2021+1/2020+1/2019-1/2018-1/2017-1/2016\ne0`)
Vậy `S={2022}`
