ko ghi lại đề 

ta thấy : 2019 - 1 = 2018 

2020 - 2 = 2018 

2021 - 3 = 2018 

2022 - 4 = 2018 

=> x = 2018

thử lại :

2018+1/2019 + 2018+2/2020 = 2018+3/2021 + 2018+4/2022

= 1 + 1 = 1 + 1

2 = 2

2020 - 2 = 2018 
2021 - 3 = 2018 
2022 - 4 = 2018 
=> x = 2018

thây zô mà thử lại

\(\frac{x+4}{2019}+\frac{x+3}{2020}=\frac{x+2}{2021}+\frac{x+1}{2020}\)

\(\Leftrightarrow(\frac{x+4}{2019}+1)+(\frac{x+3}{2020}+1)=(\frac{x+2}{2021}+1)+(\frac{x+1}{2022}+1)\)

\(\Leftrightarrow\frac{x+2023}{2019}+\frac{x+2023}{2020}=\frac{x+2023}{2021}+\frac{x+2023}{2022}\)

\(\Leftrightarrow\frac{x+2023}{2019}+\frac{x+2023}{2020}-\frac{x+2023}{2021}-\frac{x+2023}{2022}=0\)

\(\Leftrightarrow\left(x+2023\right)\left(\frac{1}{2019}+\frac{1}{2020}-\frac{1}{2021}-\frac{1}{2020}\right)=0\)

\(\Leftrightarrow x+2023=0\)

\(\Leftrightarrow x=-2023\)

Nhầm đề :( Với bước thứ 4 sửa thành ( 1/2019 + 1/2020 - 1/2021 - 1/2022 ) 

\(\frac{x+4}{2019}+\frac{x+3}{2020}=\frac{x+2}{2021}+\frac{x+1}{2022}\)

\(\Leftrightarrow\)\(\frac{x+4}{2019}+\frac{x+3}{2020}+2=\frac{x+2}{2021}+\frac{x+1}{2022}+2\)

\(\Leftrightarrow\)\(\left(\frac{x+4}{2019}+1\right)+\left(\frac{x+3}{2020}+1\right)=\left(\frac{x+2}{2021}+1\right)+\left(\frac{x+1}{2022}+1\right)\)

\(\Leftrightarrow\)\(\left(\frac{x+4}{2019}+\frac{2019}{2019}\right)+\left(\frac{x+3}{2020}+\frac{2020}{2020}\right)\)\(=\)\(\left(\frac{x+2}{2021}+\frac{2021}{2021}\right)+\left(\frac{x+1}{2022}+\frac{2022}{2022}\right)\)

\(\Leftrightarrow\)\(\frac{x+2023}{2019}+\frac{x+2023}{2020}=\frac{x+2023}{2021}+\frac{2023}{2022}\)

\(\Leftrightarrow\)\(\frac{x+2023}{2019}+\frac{x+2023}{2020}-\frac{x+2023}{2021}-\frac{x+2023}{2022}=0\)

\(\Leftrightarrow\) \(\left(x+2023\right).\left(\frac{1}{2019}+\frac{1}{2020}-\frac{1}{2021}-\frac{1}{2022}\right)=0\)

\(\Leftrightarrow\)\(x+2023=0\)      ( Vì \(\frac{1}{2019}+\frac{1}{2020}-\frac{1}{2021}-\frac{1}{2022}\ne0\))

\(\Leftrightarrow\)\(x=-2023\)

Vậy x = -2023

\(\frac{x-4}{2021}+\frac{x-3}{2020}=\frac{x-2}{2019}+\frac{x-1}{2018}\)

\(\Leftrightarrow\left(\frac{x-4}{2021}+1\right)+\left(\frac{x-3}{2020}+1\right)=\left(\frac{x-2}{2019}+1\right)+\left(\frac{x-1}{2018}+1\right)\)

\(\Leftrightarrow\frac{x+2017}{2021}+\frac{x+2017}{2020}=\frac{x+2017}{2019}+\frac{x+2017}{2018}\)

\(\Leftrightarrow\frac{x+2017}{2021}+\frac{x+2017}{2020}-\frac{x+2017}{2019}-\frac{x+2017}{2018}=0\)

\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2021}+\frac{1}{2020}-\frac{1}{2019}-\frac{1}{2018}\right)=0\)

Mà \(\left(\frac{1}{2021}+\frac{1}{2020}-\frac{1}{2019}-\frac{1}{2018}\right)\ne0\)

\(\Leftrightarrow x+2017=0\)

\(\Leftrightarrow x=-2017\)

Vậy ..

=> (x-4/2021 +1) + (x-3/2020 +1) = (x-2/2019 +1)+ (x-1/2018 +1)

=> x+2017/2021 + x+2017/2020 = x+2017/2019 + x+2017/2018

=> x+2017/2018 + x+2017/2018 - x+2017/2020 - x+2017/2021 = 0

=> (x+2017).(1/2018+1/2019+1/2020+1/2021) = 0

=> x+2017 = 0 ( vì 1/2018+1/2019+1/2020+1/2021 > 0 )

=> x=-2017

Vậy x=-2017

k mk nha

\(\frac{x-4}{2021}+\frac{x-3}{2020}=\frac{x-2}{2019}+\frac{x-1}{2018}\)

\(\left(\frac{x-4}{2021}+1\right)+\left(\frac{x-3}{2020}+1\right)=\left(\frac{x-2}{2019}+1\right)+\left(\frac{x-1}{2018}+1\right)\)

\(\frac{x-2017}{2021}+\frac{x-2017}{2020}=\frac{x-2017}{2019}+\frac{x-2017}{2018}\)

\(\frac{x-2017}{2021}+\frac{x-2017}{2020}-\frac{x-2017}{2019}-\frac{x-2017}{2018}=0\)

\(\left(x-2017\right).\left(\frac{1}{2021}+\frac{1}{2020}-\frac{1}{2019}-\frac{1}{2018}\right)=0\)

vì \(\frac{1}{2021}+\frac{1}{2020}-\frac{1}{2019}-\frac{1}{2018}\ne0\)nên x - 2017 = 0 \(\Rightarrow\)x = 2017

\(\dfrac{x-4}{2022}+\dfrac{x-3}{2021}+\dfrac{x-2}{2020}+\dfrac{x-1}{2019}\text{=}-4\)

\(\dfrac{x-4}{2022}+\dfrac{x-3}{2021}+\dfrac{x-2}{2020}+\dfrac{x-1}{2019}+4\text{=}0\)

\(\left(\dfrac{x-4}{2022}+1\right)+\left(\dfrac{x-3}{2021}+1\right)+\left(\dfrac{x-2}{2020}+1\right)+\left(\dfrac{x-1}{2019}+1\right)\text{=}0\)

\(\dfrac{x-2018}{2022}+\dfrac{x-2018}{2021}+\dfrac{x-2018}{2020}+\dfrac{x-2018}{2019}\text{=}0\)

\(\left(x-2018\right)\left(\dfrac{1}{2022}+\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}\right)\text{=}0\)

\(Do:\) \(\dfrac{1}{2022}+\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}\ne0\)

\(x-2018\text{=}0\)

\(x\text{=}2018\)

\(Vậy...\)

refer

https://lazi.vn/edu/exercise/634984/tim-x-biet-x-1-2019-x-2-2020-x-3-2021x-4-2022

\(\frac{x+1}{2018}+\frac{x+2}{2019}=\frac{x+3}{2020}+\frac{x+4}{2021}\)

\(\Leftrightarrow\left(\frac{x+1}{2018}-1\right)+\left(\frac{x+2}{2019}-1\right)=\left(\frac{x+3}{2020}-1\right)+\left(\frac{x+4}{2021}-1\right)\)

\(\Leftrightarrow\frac{x-2017}{2018}+\frac{x-2017}{2019}=\frac{x-2017}{2020}+\frac{x-2017}{2021}\)

\(\Leftrightarrow\left(x-2017\right)\left(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021}\right)=0\)

\(\Leftrightarrow x-2017=0\)\(\left(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021}\ne0\right)\)

\(\Leftrightarrow x=2017\)

Vậy \(S=\left\{2017\right\}\)

\(\frac{x+1}{2018}+\frac{x+1}{2019}=\frac{x+1}{2020}+\frac{x+1}{2021}\Leftrightarrow\frac{x+1}{2018}+\frac{x+1}{2019}-\frac{x+1}{2020}-\frac{x+1}{2021}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021}\right)=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

KL: ................

\(\Rightarrow2019\left|x-1\right|+2020\left|y-2\right|+2021\left|y-3\right|+2022\left|y-4\right|=2020+2022\)

\(\Rightarrow\hept{\begin{cases}\left|y-2\right|=1\\\left|x-1\right|=0\\\left|y-4\right|=1\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=3\end{cases}}}\)

Tổng = 4042 nha !!!