\(.c.\)
Ta có : \(A=\left|x+2\right|+\left|5-x\right|\ge\)\(\left|x+2+5-x\right|\)\(=7\)
Dấu '' = '' xảy ra \(\Leftrightarrow\left(x+2\right).\left(5-x\right)\ge0\)
\(\Leftrightarrow\left[\begin{matrix}\left\{\begin{matrix}\left(x+2\right)\ge0\\\left(5-x\right)\ge0\end{matrix}\right.\\\left\{\begin{matrix}\left(x+2\right)\le0\\\left(5-x\right)\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}\left\{\begin{matrix}x\ge-2\\x\le5\end{matrix}\right.\\\left\{\begin{matrix}x\le-2\\x\ge5\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow-2\le x\le5\)
Vậy : Giá trị nhỏ nhất của \(A=7\Leftrightarrow-2\le x\le5\)
