Question

[xung223]

(x+1)/2021+(x+2)/2020+(x+3)/2019+(x+2028)/2=0

Answer 1

[⭐Hannie⭐]

thêm \(\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}+\dfrac{1}{2}\ne0\) nên nó z nha :Đ giải thích ấy

Answer 2

[Ngô Hải Nam]

`<=>(x+1)/2021+1+(x+2)/2020+1+(x+3)/2019+1+(x+2028)/2-3=0`

`<=>(x+2022)/2021+(x+2022)/2020+(x+2022)/2019+(x+2022)/2=0`

`<=>(x+2022)(1/2021+1/2020+1/2019+1/2)=0`

`<=>x+2022=0`

`<=>x=-2022`