x + 1 2020 + x + 2 2019 + x + 3 2018=-3

Giải thích các bước giải:

`(x+1)/(2020)+(x+2)/(2019)+(x+3)/(2018)=-3`

`<=>(x+1)/(2020)+(x+2)/(2019)+(x+3)/(2018)+3=0`

`<=>((x+1)/(2020)+1)+((x+2)/(2019)+1)+((x+3)/(2018)+1)=0`

`<=>(x+1+2020)/(2020)+(x+2+2019)/(2019)+(x+3+2018)/(2018)=0`

`<=>(x+2021)/(2020)+(x+2021)/(2019)+(x+2021)/(2018)=0`

`<=>(x+2021)(1/(2020)+1/(2019)+1/(2018))=0`

`<=>x+2021=0` `(\text{vì}: 1/(2020)+1/(2019)+1/(2018) >0)`

`<=>x=-2021`

`\text{Vậy phương trình có tập nghiệm: S}={-2021}`

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

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

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

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

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

⇒ \(x+2021=0\)

⇔ \(x=-2021\)

\(Vậy\) \(x=-2021\)