Ta có:\(\frac{x-2}{2017}+1+\frac{x-3}{2018}+1=\frac{x-4}{2019}+1+\frac{x-5}{2020}+1\)

\(\Rightarrow\frac{x+2015}{2017}+\frac{x+2015}{2018}-\frac{x+2015}{2019}-\frac{x+2015}{2020}=0\)

\(\Rightarrow\left(x+2015\right)\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)

Mà \(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}>0\)

\(\Rightarrow x+2015=0\Rightarrow x=-2015\)

\(S=\left\{-2015\right\}\)

gợi ý 

2017-x-2=2018-3-x=2019-4-x=2020-5-x

\(\frac{x-2}{2017}+\frac{x-3}{2018}=\frac{x-4}{2019}+\frac{x-5}{2020}\)

\(\Leftrightarrow\left(\frac{x-2}{2017}+1\right)+\left(\frac{x-3}{2018}+1\right)=\left(\frac{x-4}{2019}+1\right)+\left(\frac{x-5}{2020}+1\right)\)

\(\Leftrightarrow\frac{x-2+2017}{2017}+\frac{x-3+2018}{2018}=\frac{x-4+2019}{2019}+\frac{x-5+2020}{2020}\)

\(\Leftrightarrow\frac{x+2015}{2017}+\frac{x+2015}{2018}-\frac{x+2015}{2019}-\frac{x+2015}{2020}=0\)

\(\Leftrightarrow\left(x+2015\right)\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)

\(\Leftrightarrow x+2015=0\)

\(\Leftrightarrow x=-2015\)

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

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

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

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

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

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

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

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

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

a, Làm

\(\frac{x+1}{2020}+\frac{x+2}{2019}+\frac{x+3}{2018}=\frac{x+4}{2017}+\frac{x+5}{2016}+\frac{x+6}{2015}\)

<=>\(\frac{x+2021}{2020}+\frac{x+2021}{2019}+\frac{x+2021}{2018}=\frac{x+2021}{2017}+\frac{x+2021}{2016}+\frac{x+2021}{2015}\)

<=>\(\left(x+2021\right)\left(\frac{1}{2020}+\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\right)=0\)

<=> x+2021=0

<=> x=-2021

Kl:......................

b, Làmmmmm

\(\frac{2-x}{2004}-1=\frac{1-x}{2005}-\frac{x}{2006}\)

<=> \(\frac{2006-x}{2004}=\frac{2006-x}{2005}+\frac{2006-x}{2006}\)

<=> \(\left(2006-x\right)\left(\frac{1}{2004}-\frac{1}{2005}-\frac{1}{2006}\right)=0< =>2006-x=0\)

<=> x=2006

Kl:..............

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

Gợi ý :

Bài 1 : Cộng thêm 1 vào 3 phân thức đầu, trừ cho 3 ở phân thức thứ 4, có nhân tử chung là (x+2020)

Bài 2 : Trừ mỗi phân thức cho 1, chuyển vế và có nhân tử chung là (x-2021)

Bài 3 : Phân thức thứ nhất trừ đi 1, phân thức hai trù đi 2, phân thức ba trừ đi 3, phân thức bốn trừ cho 4, phân thức 5 trừ cho 5. Có nhân tử chung là (x-100)

bài 3

\(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15.\)

=>\(\frac{x-90}{10}-1+\frac{x-76}{12}-2+\frac{x-58}{14}-3+\frac{x-36}{16}-4+\frac{x-15}{17}-5=0\)

=>\(\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)

=>\(\left(x-100\right).\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)

=>(x-100)=0 do \(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\)

=> x=100

\(\frac{x+2}{2018}+\frac{x+3}{2017}+\frac{x+4}{2016}+\frac{x+2036}{6}=0\)

\(\Leftrightarrow\frac{x+2}{2018}+1+\frac{x+3}{2017}+1+\frac{x+4}{2016}+1+\frac{x+2038}{6}-3=0\)

\(\Leftrightarrow\frac{x+2020}{2018}+\frac{x+2020}{2017}+\frac{x+2020}{2016}+\frac{x+2020}{6}=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}+\frac{1}{6}\right)=0\)

có : \(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}+\frac{1}{6}\ne0\)

\(\Leftrightarrow x+2020=0\)

\(\Leftrightarrow x=-2020\)

\(\frac{x-3}{2018}+\frac{x-2}{2019}=\frac{x-2019}{2}+\frac{x-2018}{3}\)

\(\Leftrightarrow\frac{x-3}{2018}-1+\frac{x-2}{2019}-1=\frac{x-2019}{2}-1+\frac{x-2018}{3}-1\)

\(\Leftrightarrow\frac{x-2021}{2018}+\frac{x-2021}{2019}=\frac{x-2021}{2}+\frac{x-2021}{3}\)

bài 3 thì lần lượt trừ đi 1; 2; 3; 4; 5