Đáp án:
$a)x=-4\\ b) x=-2023$
Giải thích các bước giải:
$a)\dfrac{x+4}{2000}+\dfrac{x+4}{2021}=\dfrac{x+4}{2020}+\dfrac{x+4}{2022}\\ \Leftrightarrow \dfrac{x+4}{2000}+\dfrac{x+4}{2021}-\dfrac{x+4}{2020}-\dfrac{x+4}{2022}=0\\ \Leftrightarrow (x+4)\underbrace{\left(\dfrac{1}{2000}+\dfrac{1}{2021}-\dfrac{1}{2020}-\dfrac{1}{2022}\right)}_{\ne 0}=0\\ \Leftrightarrow x+4=0\\ \Leftrightarrow x=-4\\ b)\dfrac{x+4}{2019}+\dfrac{x+3}{2020}=\dfrac{x+2}{2021}+\dfrac{x+1}{2022} \\ \Leftrightarrow \dfrac{x+4}{2019}+\dfrac{x+3}{2020}-\dfrac{x+2}{2021}-\dfrac{x+1}{2022}=0\\ \Leftrightarrow \dfrac{x+4}{2019}+1+\dfrac{x+3}{2020}+1-\dfrac{x+2}{2021}-1-\dfrac{x+1}{2022}-1=0\\ \Leftrightarrow \dfrac{x+4}{2019}+1+\dfrac{x+3}{2020}+1-\left(\dfrac{x+2}{2021}+1\right)-\left(\dfrac{x+1}{2022}+1\right)=0\\ \Leftrightarrow \dfrac{x+4+2019}{2019}+\dfrac{x+3+2020}{2020}-\dfrac{x+2+2021}{2021}-\dfrac{x+1+2022}{2022}=0\\ \Leftrightarrow \dfrac{x+2023}{2019}+\dfrac{x+2023}{2020}-\dfrac{x+2023}{2021}-\dfrac{x+2023}{2022}=0\\ \Leftrightarrow (x+2023) \underbrace{\left(\dfrac{1}{2019}+\dfrac{1}{2020}-\dfrac{1}{2021}-\dfrac{1}{2022}\right)}_{\ne 0}=0\\ \Leftrightarrow x+2023=0\\ \Leftrightarrow x=-2023$
