Giải phương trình:(x-3/2019)+(x-2/2020)+(x-1/2021)=3

$\text{$\dfrac{x - 3}{2019}$ + $\dfrac{x - 2}{2020}$ + $\dfrac{x - 1}{2021}$ = 3}$

$\text{⇔ $\dfrac{x - 3}{2019}$ - 1 + $\dfrac{x - 2}{2020}$ - 1 + $\dfrac{x - 1}{2021}$ - 1 = 3 - 1 - 1 - 1}$

$\text{⇔ ($\dfrac{x - 3}{2019}$ - 1) + ($\dfrac{x - 2}{2020}$ - 1) + ($\dfrac{x - 1}{2021}$ - 1) = 0}$

$\text{⇔ $\dfrac{x - 3 - 2019}{2019}$ + $\dfrac{x - 2 - 2020}{2020}$ + $\dfrac{x - 1 - 2021}{2021}$ = 0}$

$\text{⇔ $\dfrac{x - 2022}{2019}$ + $\dfrac{x - 2022}{2020}$ + $\dfrac{x - 2022}{2021}$ = 0}$

$\text{⇔ (x - 2022)($\dfrac{1}{2019}$ + $\dfrac{1}{2020}$ + $\dfrac{1}{2021}$) = 0}$

$\text{⇔ x - 2022 = 0}$

$\text{⇔ x = 2022}$

$\text{Vậy phương trình có tập nghiệm là S = {2022}}$

$\textit{Ha1zzz}$