Đáp án:
` S = {2016} `
Lời giải chi tiết:
` x/2016 + \frac{x+1}{2017} + \frac{x+2}{2018} + \frac{x+3}{2019} + \frac{x + 4}{2020} = 5 `
` <=> (x/2016 + \frac{x+1}{2017} + \frac{x+2}{2018} + \frac{x+3}{2019} + \frac{x + 4}{2020})-5 = 0 `
` <=> x/2016 - 1 + \frac{x+1}{2017} - 1 + \frac{x+2}{2018} - 1 + \frac{x+3}{2019} - 1 + \frac{x + 4}{2020} - 1 = 0 `
` <=> \frac{x-2016}{2016} + \frac{x-2016}{2017} + \frac{x-2016}{2018} + \frac{x-2016}{2019} + \frac{x-2016}{2020} = 0 `
` <=> (x - 2016)(1/2016 + 1/2017 + 1/2018 + 1/2019 + 1/2020) = 0 `
` <=> x - 2016 = 0 `
` <=> x = 2016 `
Vậy ` S = {2016} `
