Đề: \(\frac{x-2}{2020}+\frac{x-3}{2019}=\frac{x-4}{2018}+\frac{x-5}{2017}\)
⇔ \(\left(\frac{x-2}{2020}-1\right)+\left(\frac{x-3}{2019}-1\right)=\left(\frac{x-4}{2018}-1\right)+\left(\frac{x-5}{2017}-1\right)\)
⇔ \(\frac{x-2022}{2020}+\frac{x-2022}{2019}=\frac{x-2022}{2018}+\frac{x-2022}{2017}\)
⇔\(\frac{x-2022}{2020}+\frac{x-2022}{2019}-\frac{x-2022}{2018}-\frac{x-2022}{2017}=0\)
⇔ \(\left(x-2022\right)\)\(\left(\frac{1}{2020}+\frac{1}{2019}-\frac{1}{2018}-\frac{1}{2017}\right)\) = 0
Nên x - 2022 = 0 ⇔ x = 2022
Mà \(\left(\frac{1}{2020}+\frac{1}{2019}-\frac{1}{2018}-\frac{1}{2017}\right)\)≠0
Vậy nghiệm của pt là x = 2022
