a, \(\dfrac{2017.2021-4031}{2020+2017.2018}\)

= \(\dfrac{2017\left(2018+3\right)-4031}{2020+2017.2018}\)

= \(\dfrac{2017.2018+2017.3-4031}{2020+2017.2018}\)

= \(\dfrac{2017.2018+2020}{2020+2017.2018}\)

= 1
@Nguyen Thi Ngoc Linh

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

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

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

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

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

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

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

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

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

\(\Leftrightarrow\frac{x+1+2019}{2019}+\frac{x+2+2018}{2018}+\frac{x+3+2017}{2017}=\frac{x-1+2021}{2021}+\frac{x-2+2022}{2022}+\frac{x-3+2023}{2023}\)\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}+\frac{x+2020}{2017}=\frac{x+2020}{2021}+\frac{x+2020}{2022}+\frac{x+2020}{2023}\)

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

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

\(\Leftrightarrow x+2020=0\)

\(\Leftrightarrow x=-2020\)

\(\dfrac{x+2017}{x+2018}=\dfrac{2020}{2021}\)

\(\Leftrightarrow1-\dfrac{x+2017}{x+2018}=1-\dfrac{2020}{2021}\)

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

\(\Leftrightarrow\dfrac{\left(x+2018\right)-\left(x+2017\right)}{x+2018}=\dfrac{2021-2020}{2021}\)

\(\Leftrightarrow\dfrac{x+2018-x-2017}{x+2018}=\dfrac{1}{2021}\)

\(\Leftrightarrow\dfrac{\left(2018-2017\right)+\left(x+x\right)}{x+2018}=\dfrac{1}{2021}\)

\(\Leftrightarrow\dfrac{1}{x+2018}=\dfrac{1}{2021}\)

\(\Leftrightarrow x+2018=2021\)

\(\Leftrightarrow x=3\left(tm\right)\)

vậy ....

khó hiểu vcl

đúng lun ko hiểu một chút nào
 

mãi mới có người đồng cảm...T-T

\(\dfrac{x-1}{2021}+\dfrac{x-2}{2020}=\dfrac{x-5}{2017}+\dfrac{x-7}{2015}\\ \dfrac{x-1}{2021}+\dfrac{x-2}{2020}-2=\dfrac{x-5}{2017}+\dfrac{x-7}{2015}-2\\ \dfrac{x-1}{2021}+\dfrac{x-2}{2020}-1-1=\dfrac{x-5}{2017}+\dfrac{x-7}{2015}-1-1\\\left(\dfrac{x-1}{2021}-1\right)+\left(\dfrac{x-2}{2020}-1\right)=\left(\dfrac{x-5}{2017}-1\right)+\left(\dfrac{x-7}{2015}-1\right)\\ \dfrac{x-2022}{2021}+\dfrac{x-2022}{2020}=\dfrac{x-2022}{2017}+\dfrac{x-2022}{2015}\\ \dfrac{x-2022}{2021}+\dfrac{x-2022}{2020}-\dfrac{x-2022}{2017}-\dfrac{x-2022}{2015}=0\\ \left(x-2022\right)\left(\dfrac{1}{2021}+\dfrac{1}{2020}-\dfrac{1}{2017}-\dfrac{1}{2015}\right)=0\)

mà `(1/2021+1/2020-1/2017-1/2015 \ne 0`

nên `x-2022=0`

`x=2022`

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

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

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

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

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

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

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

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

\(x=0-2022\) 

\(x=-2022\)

\(\frac{x+1}{2020}+\frac{x+2}{2019}+\frac{x+3}{2018}+\frac{x+4}{2017}=-4\)

=> \(\left[\frac{x+1}{2020}+1\right]+\left[\frac{x+2}{2019}+1\right]+\left[\frac{x+3}{2018}+1\right]+\left[\frac{x+4}{2017}+1\right]=-4\)

=> \(\left[\frac{x+1}{2020}+\frac{2020}{2020}\right]+\left[\frac{x+2}{2019}+\frac{2019}{2019}\right]+\left[\frac{x+3}{2018}+\frac{2018}{2018}\right]+\left[\frac{x+4}{2017}+\frac{2017}{2017}\right]=-4\)

=>  \(\frac{x+2021}{2020}+\frac{x+2021}{2019}+\frac{x+2021}{2018}+\frac{x+2021}{2017}=-4\)

=> \(\left[x+2021\right]\left[\frac{1}{2000}+\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}\right]=-4\)

Do \(\frac{1}{2020}>\frac{1}{2019}>\frac{1}{2018}>\frac{1}{2017}\)nên \(\frac{1}{2000}+\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}\ne0\)

Do đó : x + 2021 = -4 => x = -4 - 2021 = -2025