#PLPT
Đáp án+Giải thích các bước giải:
$\frac{x-4}{2019}$ `+` $\frac{x-3}{2020}$ `=` $\frac{x-2}{2021}$ `+` $\frac{x-1}{2022}$
⇔$\frac{x-4}{2019}$ `+` $\frac{x-3}{2020}$`-2=` $\frac{x-2}{2021}$ `+` $\frac{x-1}{2022}$`-2`
⇔$\frac{x-4}{2019}$ `+` $\frac{x-3}{2020}$`-1-1=` $\frac{x-2}{2021}$ + $\frac{x-1}{2022}$`-1-1`
⇔$(\frac{x-4}{2019}$`-1`)`+`($\frac{x-3}{2020}$`-1`)`=` ($\frac{x-2}{2021}$`-1`) + ($\frac{x-1}{2022}$`-1`)
⇔$(\frac{x-4}{2019}$`-`$\frac{2019}{2019}$ )`+`($\frac{x-3}{2020}$`-`$\frac{2020}{2020}$ )`=`($\frac{x-2}{2021}$-$\frac{2021}{2021}$ ) `+` ($\frac{x-1}{2022}$-$\frac{2022}{2022}$ )
⇔$\frac{x-4-2019}{2019}$ `+`$\frac{x-3-2020}{2020}$ `=` $\frac{x-2-2021}{2021}$ `+` $\frac{x-1-2022}{2022}$
⇔$\frac{x-2023}{2019}$ `+` $\frac{x-2023}{2020}$`=` $\frac{x-2023}{2021}$ `+`$\frac{x-2023}{2022}$
⇔$\frac{x-2023}{2019}$ `+` $\frac{x-2023}{2020}$`-`$\frac{x-2023}{2021}$-$\frac{x-2023}{2022}$` =0`
⇔`(x-2023)`($\frac{1}{2019}$ + $\frac{1}{2020}$ - $\frac{1}{2021}$ - $\frac{1}{2022}$ )`=0`
⇔`x-2023=0`(vì $\frac{1}{2019}$ + $\frac{1}{2020}$ - $\frac{1}{2021}$ - $\frac{1}{2022}$$\neq$ `0`)
⇔`x=2023`
Vậy phương trình có tập nghiệm là $S$`={2023}`