\(Q=\left|x-2020\right|+\left|x-2021\right|+\left|x-2022\right|\)
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Cho đa thức \(f\left(x\right)\)bậc 3 với hệ số \(x^3\)là số nguyên dương thỏa mãn:
\(f\left(2019\right)=2020;f\left(2020\right)=2021\)
CMR \(f\left(2021\right)-f\left(2018\right)\)là hợp số
\(=\dfrac{\left|x-2020\right|+2022-1}{\left|x-2020\right|+2022}=1-\dfrac{1}{\left|x-2020\right|+2022}\\ mà\left|x-2020\right|\ge0\\ \Rightarrow\left|x-2022\right|+2022\ge2022\)
\(\Rightarrow\dfrac{1}{\left|x-2020\right|+2022}\le\dfrac{1}{2022}\\ =1-\dfrac{1}{\left|x-2020\right|+2022}\ge1-\dfrac{1}{2022}\\ =\dfrac{2021}{2022}\\ \Rightarrow B_{min}=\dfrac{2021}{2022}.tại.x-2020=0\Rightarrow x=2020\)
a)
`(2x-1)(x+2/3)=0`
\(< =>\left[{}\begin{matrix}2x-1=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
b)
\(\dfrac{x+4}{2019}+\dfrac{x+3}{2020}=\dfrac{x+2}{2021}+\dfrac{x+1}{2022}\)
\(< =>\dfrac{x+4}{2019}+1+\dfrac{x+3}{2020}+1=\dfrac{x+2}{2021}+1+\dfrac{x+1}{2022}+1\)
\(< =>\dfrac{x+2023}{2019}+\dfrac{x+2023}{2020}=\dfrac{x+2023}{2021}+\dfrac{x+2023}{2022}\)
\(< =>\left(x+2023\right)\left(\dfrac{1}{2019}+\dfrac{1}{2020}-\dfrac{1}{2021}-\dfrac{1}{2022}\right)=0\)
\(< =>x+2023=0\left(\dfrac{1}{2019}+\dfrac{1}{2020}-\dfrac{1}{2021}-\dfrac{1}{2022}\ne0\right)\\ < =>x=-2023\)
đề bài yêu cầu gì bn?
chắc rút gọn ash bn.