\(\dfrac{x+1}{2020}+\dfrac{x+2}{2019}+\dfrac{x+3}{2018}+\dfrac{x+4}{2017}+4=0\)

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

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

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

\(Do\) \(\left(\dfrac{1}{2020}+\dfrac{1}{2019}+\dfrac{1}{2018}+\dfrac{1}{2017}\right)\ne0\)

⇒ \(x+2021=0\)

⇔ \(x=-2021\)

\(Vậy\) \(x=-2021\)

(x+4)/2017 + (x+3)/2018 = (x+2)/2019 + (x+1)/2020

=> (x+4)/2017 + 1 + (x+3)/2018 + 1 = (x + 2)/2019 + 1 + (x + 1)/2020 + 1

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

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

mà 1/2017 + 1/2018 khác 1/2019 + 1/2020

=> x + 2021 = 0

=> x = -2021

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

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

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

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

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

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

Vì \(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\ne0\)

\(\Rightarrow x-2021=0\)

Vậy \(x=2021\)

ko ghi lại đề

=>x+2021=0

x=-2021

hc tốt

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

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

\(\Rightarrow1+\frac{x-1}{2017}+1+\frac{x-2}{2018}=1+\frac{x-4}{2020}+1+\frac{x-3}{2019}\)

\(\Rightarrow\frac{2016+x}{2017}+\frac{2016+x}{2018}=\frac{2016+x}{2020}+\frac{2016+x}{2019}\)

\(\Rightarrow\frac{2016+x}{2017}+\frac{2016+x}{2018}-\frac{2016+x}{2019}-\frac{2016+x}{2020}=0\)

\(\Rightarrow\left(2016+x\right)\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)
\(\text{Mà : }\)\(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\ne0\)

\(\text{Nên : }\) \(2016+x=0\)

\(\Rightarrow x=-2016\)

Giỏi wá!!!!!!!!

Ta có : x−12017 +x−22018 −x−32019 =x−42020 

⇒x−12017 +x−22018 =x−42020 +x−32019 

⇒1+x−12017 +1+x−22018 =1+x−42020 +1+x−32019 

⇒2016+x2017 +2016+x2018 =2016+x2020 +2016+x2019 

⇒2016+x2017 +2016+x2018 −2016+x2019 −2016+x2020 =0

⇒(2016+x)(12017 +12018 −12019 −12020 )=0
Mà : 12017 +12018 −12019 −12020 ≠0

Nên : 2016+x=0

⇒x=−2016

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

\(\Leftrightarrow x+1=0\)

hay x=-1

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

<=> \(\frac{x-2}{2017}+1+\frac{x-3}{2018}+1=\frac{x-4}{2019}+1+\frac{x-5}{2020}+1\)

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

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

<=> x + 2015 = 0  ( vì \(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\ne0\)) 

<=> x = - 2015 

Vậy x = -2015.

Giải phương trình :

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

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

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

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

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

\(\Rightarrow x+2015=0\)

\(\Rightarrow x=-2015\)

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

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

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

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

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

\(\Rightarrow\left(x+15\right)\left(\frac{1}{2017}+\frac{1}{2018}-\frac{1}{2019}-\frac{1}{2020}\right)=0\)

\(\Rightarrow x+2015=0\)

\(\Rightarrow x=-2015\)

Vậy x = - 2015 

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

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

⇔ \(\frac{x-2022}{2020}+\frac{x-2022}{2019}=\frac{x-2022}{2018}+\frac{x-2022}{2017}\)

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

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

Nên x - 2022 = 0 ⇔ x = 2022

Mà \(\left(\frac{1}{2020}+\frac{1}{2019}-\frac{1}{2018}-\frac{1}{2017}\right)\)≠0

Vậy nghiệm của pt là x = 2022

\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}=\dfrac{x+1}{6}\)

\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}-\dfrac{x+1}{6}=0\)

\(\left(x+1\right)\left(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\right)=0\)

\(\)vì \(\dfrac{1}{3}>\dfrac{1}{6};\dfrac{1}{4}>\dfrac{1}{6};\dfrac{1}{5}>\dfrac{1}{6}=>\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}>0\)

\(=>x+1=0\)

\(=>x=-1\)

b,

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

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

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

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

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

Vì \(\dfrac{1}{2020}< \dfrac{1}{2018};\dfrac{1}{2019}< \dfrac{1}{2017}=>\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}< 0\)

\(=>x+2021=0\)

\(=>x=-2021\)

c,

\(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)

\(\left(\dfrac{x+2}{327}+1\right)+\left(\dfrac{x+3}{326}+1\right)+\left(\dfrac{x+4}{325}+1\right)+\left(\dfrac{x+5}{324}+1\right)+\left(\dfrac{x+349}{5}-4\right)=0\)

\(\dfrac{x+329}{327}+\dfrac{x+329}{326}+\dfrac{x+329}{325}+\dfrac{x+329}{324}+\dfrac{x+329}{5}=0\)

\(=>\left(x+329\right)\left(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}\right)=0\)

Vì \(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}>0\)

\(=>x+329=0\)

\(=>x=-329\)