Ta có: `A=|x-2|+|x-3|+|x-4|+|x-5|`
`=|x-2|+|x-3|+|4-x|+|5-x|`
`≥x-2+x-3+4-x+5-x=4`
Dấu `"="` xảy ra khi $\begin{cases}x-2\ge0\\x-3\ge0\\4-x\ge0\\5-x\ge0\end{cases}\Rightarrow\begin{cases}x\ge1\\x\ge2\\x\le3\\x\le4\end{cases}$
$\Rightarrow3\le x\le4$
Vậy $Min_{A}=4$ khi `3≤x≤4`