\(y\left(0\right)=\left|m-2\right|\) ; \(y\left(1\right)=\left|m-3\right|\) ; \(y\left(4\right)=\left|m+6\right|\)
\(\Rightarrow y_{max}=max\left\{\left|m-3\right|;\left|m+6\right|\right\}\)
TH1: \(\left\{{}\begin{matrix}\left|m-3\right|\ge\left|m+6\right|\\\left|m-3\right|=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m\le-\frac{3}{2}\\\left[{}\begin{matrix}m=13\\m=-7\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow m=-7\)
TH2: \(\left\{{}\begin{matrix}\left|m+6\right|\ge\left|m-3\right|\\\left|m+6\right|=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m\ge-\frac{3}{2}\\\left[{}\begin{matrix}m=4\\m=-16\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow m=4\)
Vậy \(m=\left\{-7;4\right\}\)
