Ta có: \(Q=\left|x-2\right|+\left|x-8\right|=\left|x-2\right|+\left|8-x\right|\)
Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(Q=\left|x-2\right|+\left|8-x\right|\ge\left|x-2+8-x\right|=\left|6\right|=6\)
Dấu " = " khi \(\left\{{}\begin{matrix}x-2\ge0\\8-x\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge2\\x\le8\end{matrix}\right.\Rightarrow2\le x\le8\)
Mà \(x\in Z\)
\(\Rightarrow x\in\left\{2;3;4;5;6;7;8\right\}\)
Vậy \(MIN_Q=6\) khi \(x\in\left\{2;3;4;5;6;7;8\right\}\)
Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(Q=\left|x-2\right|+\left|x-8\right|\)
\(=\left|x-2\right|+\left|8-x\right|\)
\(\ge\left|x-2+8-x\right|=6\)
Đẳng thức xảy ra khi \(2\le x\le8\)
