Đặt \(A=\left|2022-x\right|+\left|x-2020\right|\)
\(\Rightarrow A\ge\left|2022-x+x-2020\right|=2\)
\(A_{min}=2\) khi \(\left(2022-x\right)\left(x-2020\right)\ge0\Rightarrow2020\le x\le2022\)
\(\left|2022-x\right|+\left|x-2020\right|\)
Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) vào biểu thức, ta được:
\(\left|2022-x\right|+\left|x-2020\right|\ge\left|2022-x+x-2020\right|=\left|2\right|=2\)
Dấu \("="\) xảy ra khi: \(\left(2022-x\right)\left(x-2020\right)\ge0\)
\(\Rightarrow\left[{}\begin{matrix}\left(2022-x\right)\left(x-2020\right)>0\\\left(2022-x\right)\left(x-2020\right)=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}2022-x>0\\x-2020>0\end{matrix}\right.\\\left\{{}\begin{matrix}2022-x< 0\\x-2020< 0\end{matrix}\right.\end{matrix}\right.\\\left[{}\begin{matrix}2022-x=0\\x-2020=0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}2022>x\\x>2020\end{matrix}\right.\\\left\{{}\begin{matrix}2022< x\\x< 2020\end{matrix}\right.\end{matrix}\right.\\\left[{}\begin{matrix}x=2022\\x=2020\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}2020< x< 2022\\2022< x< 2020\left(\text{vô lí}\right)\end{matrix}\right.\\\left[{}\begin{matrix}x=2022\\x=2020\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2020< x< 2022\\\left[{}\begin{matrix}x=2022\\x=2020\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow2020\le x\le2022\)
\(\text{#}\mathit{Toru}\)
