Ta có :
\(A=\left|x-1\right|+\left|5-x\right|\)
Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)
\(\Leftrightarrow A\ge\left|x-1+5-x\right|=\left|4\right|=4\)
Dấu "=" xảy ra khi :
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1\ge0\Leftrightarrow x\ge1\\5-x\ge0\Leftrightarrow x\le5\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\Leftrightarrow x< 1\\5-x< 0\Leftrightarrow x>5\end{matrix}\right.\end{matrix}\right.\)
Vậy ..
\(A=\left|x-1\right|+\left|5-x\right|\)
Áp dụng bđt: \( \left|a\right|+\left|b\right|\ge\left|a+b\right|\)
\(A=\left|x-1\right|+\left|5-x\right|\ge\left|x-1+5-x\right|=4\)
Dấu "=" xảy ra khi:
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1\ge0\Rightarrow x\ge1\\5-x\ge0\Rightarrow x\le5\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\Rightarrow x< 1\\5-x< 0\Rightarrow x< 5\end{matrix}\right.\end{matrix}\right.\)
Vậy \(1\le x\le5\)
