Question

 \(\dfrac{x+4}{2020}\)+ \(\dfrac{x+3}{2021}\)=\(\dfrac{x+2}{2022}\)+ \(\dfrac{x+1}{2023}\)

 

 

 

 

Answer 1

\(\dfrac{x+4}{2020}+\dfrac{x+3}{2021}=\dfrac{x+2}{2022}+\dfrac{x+1}{2023}\\ \Leftrightarrow\left(\dfrac{x+4}{2020}+1\right)+\left(\dfrac{x+3}{2021}+1\right)=\left(\dfrac{x+2}{2022}+1\right)+\left(\dfrac{x+1}{2023}+1\right)\\ \Leftrightarrow\dfrac{x+2024}{2020}+\dfrac{x+2024}{2021}-\dfrac{x+2024}{2022}-\dfrac{x+2024}{2023}=0\\ \Leftrightarrow\left(x+2024\right)\left(\dfrac{1}{2020}+\dfrac{1}{2021}-\dfrac{1}{2022}-\dfrac{1}{2023}\right)=0\\ \Leftrightarrow x+2024=0\left(vì.\dfrac{1}{2020}+\dfrac{1}{2021}-\dfrac{1}{2022}-\dfrac{1}{2023}\ne0\right)\\ \Leftrightarrow x=-2024\)

Vậy `x=-2024`