Sửa lại đề
$\dfrac{x+2}{2020}+\dfrac{x+1}{2021}-\dfrac{x-3}{2025}=-1$
$\to \dfrac{x+2}{2020}+\dfrac{x+1}{2021}-\dfrac{x-3}{2025}+1=0$
$\to (\dfrac{x+2}{2020}+1)+(\dfrac{x+1}{2021}+1)-(\dfrac{x-3}{2025}+1)=0$
$\to \dfrac{x+2+2020}{2020}+\dfrac{x+1+2021}{2021}-\dfrac{x-3+2025}{2025}=0$
$\to \dfrac{x+2022}{2020}+\dfrac{x+2022}{2021}-\dfrac{x+2022}{2025}=0$
$\to (x+2022)(\dfrac{1}{2020}+\dfrac{1}{2021}-\dfrac{1}{2025})=0$
$\text{Do:}$
$\dfrac{1}{2020}+\dfrac{1}{2021}-\dfrac{1}{2025}\ne0$
$\to x+2022=0$
$\to x=-2022$
