Đáp án:
Giải thích các bước giải:
$\frac{x +4}{2016}$ + $\frac{x+3}{2017}$ = $\frac{x+2}{2018}$ + $\frac{x+1}{2019}$
⇒(1+$\frac{x +4}{2016}$) + (1+$\frac{x+3}{2017}$) = (1+$\frac{x+2}{2018}$) + (1+$\frac{x+1}{2019}$)
⇒$\frac{x +2020}{2016}$ + $\frac{x+2020}{2017}$ = $\frac{x+2020}{2018}$ + $\frac{x+2020}{2019}$
⇒ $\frac{x +2020}{2016}$ + $\frac{x+2020}{2017}$ - $\frac{x+2020}{2018}$ - $\frac{x+2020}{2019}$ = 0
⇒ (x + 2020) . ($\frac{1}{2016}$ + $\frac{1}{2017}$ - $\frac{1}{2018}$ - $\frac{1}{2019}$) = 0
Do $\frac{1}{2016}$ + $\frac{1}{2017}$ - $\frac{1}{2018}$ - $\frac{1}{2019}$ $ \neq$ 0
⇒ x + 2020 = 0
⇒ x = -2020
Vậy x = -2020
