\(\dfrac{x-1}{2021}+\dfrac{x-2}{2020}=\dfrac{x-5}{2017}+\dfrac{x-7}{2015}\\ \dfrac{x-1}{2021}+\dfrac{x-2}{2020}-2=\dfrac{x-5}{2017}+\dfrac{x-7}{2015}-2\\ \dfrac{x-1}{2021}+\dfrac{x-2}{2020}-1-1=\dfrac{x-5}{2017}+\dfrac{x-7}{2015}-1-1\\\left(\dfrac{x-1}{2021}-1\right)+\left(\dfrac{x-2}{2020}-1\right)=\left(\dfrac{x-5}{2017}-1\right)+\left(\dfrac{x-7}{2015}-1\right)\\ \dfrac{x-2022}{2021}+\dfrac{x-2022}{2020}=\dfrac{x-2022}{2017}+\dfrac{x-2022}{2015}\\ \dfrac{x-2022}{2021}+\dfrac{x-2022}{2020}-\dfrac{x-2022}{2017}-\dfrac{x-2022}{2015}=0\\ \left(x-2022\right)\left(\dfrac{1}{2021}+\dfrac{1}{2020}-\dfrac{1}{2017}-\dfrac{1}{2015}\right)=0\)
mà `(1/2021+1/2020-1/2017-1/2015 \ne 0`
nên `x-2022=0`
`x=2022`