Question

\(A=\left|x-2001\right|+\left|x-1\right|\)

Answer 1

Ta có :

\(A=\left|x-2001\right|+\left|x-1\right|=\left|x-2001\right|+\left|1-x\right|\)

\(\Leftrightarrow A\ge\left|\left(x-2001\right)+\left(1-x\right)\right|\)

\(\Leftrightarrow A\ge\left|-2000\right|\)

\(\Leftrightarrow A\ge2000\)

Dấu "=" xảy ra khi :

\(\left(x-2001\right)\left(1-x\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2001\ge0\\1-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2001\le0\\1-x\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge2001\\1\ge x\end{matrix}\right.\\\left\{{}\begin{matrix}x\le2001\\1\le x\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2001\ge x\ge1\\x\in\varnothing\end{matrix}\right.\)

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