Mình làm đầy đủ hơn nè.
\(A=\left|x-2\right|+\left|5-x\right|\)
Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(A=\left|x-2\right|+\left|5-x\right|\ge\left|x-2+5-x\right|\)
\(\Rightarrow A\ge\left|3\right|\)
\(\Rightarrow A\ge3.\)
Dấu '' = '' xảy ra khi:
\(\left(x-2\right).\left(5-x\right)\ge0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2\ge0\\5-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2\le0\\5-x\le0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge2\\x\le5\end{matrix}\right.\\\left\{{}\begin{matrix}x\le2\\x\ge5\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2\le x\le5\\x\in\varnothing\end{matrix}\right.\)
Vậy \(MIN_A=3\) khi \(2\le x\le5.\)
Chúc bạn học tốt!
