a) Ta có: \(2\left|x-3\right|\ge0\)
\(\Rightarrow-2\left|x-3\right|\le0\)
\(\Rightarrow M=9-2\left|x-3\right|\le9\)
Dấu " = " khi \(2\left|x-3\right|=0\Rightarrow x-3=0\Rightarrow x=3\)
Vậy \(MAX_M=9\) khi x = 3
b) Ta có: \(N=\left|x-2\right|+\left|x-8\right|=\left|x-2\right|+\left|8-x\right|\)
Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(N=\left|x-2\right|+\left|8-x\right|=\left|x-2+8-x\right|=\left|-6\right|=6\)
Dấu " = " khi \(\left\{{}\begin{matrix}x-2\ge0\\8-x\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge2\\x\le8\end{matrix}\right.\Rightarrow2\le x\le8\)
Vậy \(MIN_N=6\) khi \(2\le x\le8\)
a/ \(M=9-2\left|x-3\right|\)
Ta có: \(\left|x-3\right|\ge0\forall x\)
\(\Rightarrow-2\left|x-3\right|\le0\)
Dấu ''='' xảy ra \(\Leftrightarrow x=3\)
\(\Rightarrow Max_M=9\Leftrightarrow x=3\)
b/ \(N=\left|x-2\right|+\left|x-8\right|\)
Ta có: \(\left\{{}\begin{matrix}\left|x-2\right|\ge0\forall x\\\left|x-8\right|\ge0\forall x\end{matrix}\right.\)
Dấu ''='' xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=8\end{matrix}\right.\)
\(\Rightarrow Min_N=6\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=8\end{matrix}\right.\)
