Question

x+4/2018+x+3/2019=x+2/2020+x+1/2021=4
ai làm nhanh trước mik tích

 

Answer 1

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

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

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

=>x+2022=0

=>x=-2022