\(\dfrac{x-1}{2020}=\dfrac{y-1}{2021}=\dfrac{z-1}{2022}=\dfrac{x-y}{-1}=\dfrac{y-z}{-1}=\dfrac{x-z}{-2}\)
\(\Rightarrow2\left(x-y\right)=2\left(y-z\right)=x-z\)
Vậy \(8\left(x-y\right)^2\left(y-z\right)=8\left(x-y\right)^3=\left[2\left(x-y\right)\right]^3=\left(x-z\right)^3\)
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