Đặt `B = |x - 1| + |x - 2| + |x - 3| + |x - 4|`
`= (|x - 1| + |x - 4|) + (|x - 2| + |x - 3|)`
`= (|x - 1| + |4 - x|) + (|x - 2| + |3 - x|)`
\(\Rightarrow B\ge\left|x-1+4-x\right|+\left|x-2+3-x\right|\)
\(B\ge\left|3\right|+\left|1\right|=4\)
\(\Rightarrow A\ge4+15=19\)
hay MinA = 19
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}\left(x-1\right)\left(4-x\right)\ge0\\\left(x-2\right)\left(3-x\right)\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(x-4\right)\le0\\\left(x-2\right)\left(x-3\right)\le0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}1\le x\le4\\2\le x\le3\end{matrix}\right.\Rightarrow2\le x\le3\)
Vậy MinA = 19 tại \(2\le x\le3\).
