TH1: x<2019

=>x-2019<0; x-2020<0; x-2021<0; x-2022<0

=>M=-x+2019-x+2020-x+2021-x+2022=-4x+8082

Vì hàm số M=-4x+8082 là hàm số nghịch biến trên R

nên M nhỏ nhất khi x lớn nhất

Khi x<2019 thì x không có giá trị lớn nhất

=>M không có giá trị nhỏ nhất

TH2: 2019<=x<2020

=>x-2019>=0; x-2020<0; x-2021<0; x-2022<0

=>M=x-2019-x+2020-x+2021-x+2022=-2x+4034

Vì hàm số M=-2x+4034 là hàm số nghịch biến trên R

nên M nhỏ nhất khi x lớn nhất

Khi 2019<=x<2020 thì x không có giá trị lớn nhất

=>M không có giá trị nhỏ nhất

TH3: 2020<=x<2021

=>x-2019>0; x-2020>=0; x-2021<0; x-2022<0

=>M=x-2019+x-2020+2021-x+2022-x=4(1)

TH4: 2021<=x<2022

=>x-2019>0; x-2020>0; x-2021>=0; x-2022<0

=>M=x-2019+x-2020+x-2021+2022-x=2x-4038

Vì hàm số M=2x-4038 là hàm số đồng biến trên R

nên M nhỏ nhất khi x nhỏ nhất

Với 2021<=x<2022 thì \(x_{\min}=2021\)

=>\(M_{\min}=2\cdot2021-4038=4042-4038=4\) (2)

TH5: x>=2022

=>x-2019>0; x-2020>0; x-2021>=0; x-2022>=0

=>M=x-2019+x-2020+x-2021+x-2022=4x-8082

Vì hàm số M=4x-8082 là hàm số đồng biến trên R

nên M nhỏ nhất khi x nhỏ nhất

Khi x>=2022 thì \(x_{\min}=2022\)

=>\(M_{\min}=4\cdot2022-8082=8088-8082=6\) (3)

Từ (1),(2),(3) suy ra \(M_{\min}=4\) khi 2020<=x<=2022

x=2020 nên x+1=2021

\(P\left(x\right)=x^{2021}-x^{2020}\left(x+1\right)+x^{2019}\left(x+1\right)-....+x\left(x+1\right)-2020\)

\(=x^{2021}-x^{2021}-x^{2020}+x^{2020}-...+x^2+x-2020\)

=x-2020=0

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

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

=> \(\frac{x+2022}{2018}+\frac{x+2022}{2019}=\frac{x+2022}{2020}+\frac{x+2022}{2021}\)

=> \(\frac{x+2022}{2018}+\frac{x+2022}{2019}-\frac{x+2022}{2020}-\frac{x+2022}{2021}=0\)

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

Vì \(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021}\ne0\)

=> x + 2022 = 0

=> x = -2022

Vậy x = -2022

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

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

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

\(\frac{x+2022}{2018}+\frac{x+2022}{2019}=\frac{x+2022}{2020}+\frac{x+2022}{2021}\)   

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

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

\(x+2022=0\left(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021}\ne0\right)\)   

\(x=0-2022\) 

\(x=-2022\)

\(\frac{x+1}{2019}+\frac{x+2}{2018}+\frac{x+3}{2017}=\frac{x-1}{2021}+\frac{x-2}{2022}+\frac{x-3}{2023}\)

\(\Leftrightarrow\left(\frac{x+1}{2019}+1\right)+\left(\frac{x+2}{2018}+1\right)+\left(\frac{x+3}{2017}+1\right)=\left(\frac{x-1}{2021}+1\right)+\left(\frac{x-2}{2022}+1\right)+\left(\frac{x-3}{2023}+1\right)\)

\(\Leftrightarrow\left(\frac{x+1+2019}{2019}\right)+\left(\frac{x+2+2018}{2018}\right)+\left(\frac{x+3+2017}{2017}\right)=\left(\frac{x-1+2021}{2021}\right)+\left(\frac{x-2+2022}{2022}\right)+\left(\frac{x-3+2023}{2023}\right)\)

\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}=\frac{x+2020}{2021}+\frac{x+2020}{2022}+\frac{x+2020}{2023}\)

\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}-\frac{x+2020}{2021}-\frac{x+2020}{2022}-\frac{x+2020}{2023}=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2021}-\frac{1}{2022}-\frac{1}{2023}\right)=0\)

Vì \(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2021}-\frac{1}{2022}-\frac{1}{2023}\ne0\)

=> x + 2020 = 0

=> x = -2020

            Bài làm :

Ta có :

\(\frac{x+1}{2019}+\frac{x+2}{2018}+\frac{x+3}{2017}=\frac{x-1}{2021}+\frac{x-2}{2022}+\frac{x-3}{2023}\)

\(\Leftrightarrow\left(\frac{x+1}{2019}+1\right)+\left(\frac{x+2}{2018}+1\right)+\left(\frac{x+3}{2017}+1\right)=\left(\frac{x-1}{2021}+1\right)+\left(\frac{x-2}{2022}+1\right)+\left(\frac{x-3}{2023}+1\right)\)

\(\Leftrightarrow\left(\frac{x+1+2019}{2019}\right)+\left(\frac{x+2+2018}{2018}\right)+\left(\frac{x+3+2017}{2017}\right)=\left(\frac{x-1+2021}{2021}\right)+\left(\frac{x-2+2022}{2022}\right)+\left(\frac{x-3+2023}{2023}\right)\)

\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}=\frac{x+2020}{2021}+\frac{x+2020}{2022}+\frac{x+2020}{2023}\)

\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}-\frac{x+2020}{2021}-\frac{x+2020}{2022}-\frac{x+2020}{2023}=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2021}-\frac{1}{2022}-\frac{1}{2023}\right)=0\)

 \(\text{Vì : }\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2021}-\frac{1}{2022}-\frac{1}{2023}\ne0\)

\(\Rightarrow x+2020=0\Leftrightarrow x=-2020\)

Vậy x=-2020

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

⇔ \(\dfrac{x-4}{2021}+\dfrac{x-3}{2020}-\dfrac{x-2}{2019}-\dfrac{x-1}{2018}=0\)

⇔ \(\left(1+\dfrac{x-4}{2021}\right)+\left(1+\dfrac{x-3}{2020}\right)-\left(1+\dfrac{x-2}{2019}\right)-\left(1+\dfrac{x-1}{2018}\right)=0\)⇔ \(\dfrac{x+2017}{2021}+\dfrac{x+2017}{2020}-\dfrac{x+2017}{2019}-\dfrac{x+2017}{2018}=0\)

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

⇔ x + 2017 = 0

⇔ x = -2017

Vậy x = -2017

lol

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

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

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

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

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

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

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

\(\Leftrightarrow x+2023=0\)

\(\Leftrightarrow x=-2023\)

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

Sửa đề: \(\frac{x+1}{2018}+\frac{x+1}{2019}+\frac{x+1}{2020}+\frac{x+1}{2021}=0\)

Ta có: \(\frac{x+1}{2018}+\frac{x+1}{2019}+\frac{x+1}{2020}+\frac{x+1}{2021}=0\)

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

=>x+1=0

=>x=-1

khó hiểu vcl

đúng lun ko hiểu một chút nào
 

Link bài làm của mình đây nhé 

https://olm.vn/hoi-dap/detail/831153598726.html 

day nha ban