` \frac{x+2}{2018} + \frac{x+3}{2017}=\frac{x+2018}{2} + \frac{x+2017}{2} `
` <=> \frac{x+2}{2018} + 1 + \frac{x+3}{2017} + 1 = \frac{x+2018}{2} + 1 + \frac{x+2017}{2} + 1 `
` <=> \frac{2020+x}{2018} + \frac{2020+x}{2017} = \frac{x+2020}{2} + \frac{x+2020}{3} `
` <=> (x+2020).(\frac{1}{2018}+\frac{1}{2017} - 1/2 - 1/3) = 0 `
Do: ` \frac{1}{2018} + \frac{1}{2017} - 1/2 - 1/3 ne 0 `
` <=> x+2020 = 0 `
` <=> x = -2020 `
Vậy ` S = {-2020} `
