Đáp án+Giải thích các bước giải:
`\frac{x + 2021}{2016} + \frac{x + 2021}{2017} + \frac{x + 2021}{2018} = \frac{x + 2021}{2019} + \frac{x + 2021}{2020}`
`⇔ \frac{x + 2021}{2016} + \frac{x + 2021}{2017} + \frac{x + 2021}{2018} - \frac{x + 2021}{2019} - \frac{x + 2021}{2020} = 0`
`⇔ (x + 2021) . (\frac{1}{2016} + \frac{1}{2017} + \frac{1}{2018} - \frac{1}{2019} - \frac{1}{2020}) = 0`
`⇒ x + 2021 = 0` (Vì `\frac{1}{2016} + \frac{1}{2017} + \frac{1}{2018} - \frac{1}{2019} - \frac{1}{2020} > 0`)
`⇔ x = - 2021`
