(x+1)/2017 + (x+2)/2018 = [(x+3)/2019] +1
⇔ {[(x+1)/2017] - 1} +{ [(x+2)/2018] -1} - {[(x+3)/2019] - 1} =0
⇔ (x-2016) / 2017 + ((x-2016) / 2018 - (x-2016) / 2019 = 0
⇔ (x-2016) (1/2017 + 1/2018 - 1/2019) = 0
Vì 1/2017 + 1/2018 - 1/2019 $\neq$ 0
Nên x-2016 = 0
⇔ x = 2016