\(A=\left|x-1\right|+\left|x-2\right|=\left|x-1\right|+\left|2-x\right|\)
Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(A=\left|x-1\right|+\left|2-x\right|\ge\left|x-1+2-x\right|=\left|-1\right|=1\)
Dấu " = " xảy ra khi \(\left\{{}\begin{matrix}x-1\ge0\\2-x\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge1\\x\le2\end{matrix}\right.\Rightarrow1\le x\le2\)
Vì \(x\in Z\Rightarrow x\in\left\{1;2\right\}\)
Vậy \(MIN_A=1\) khi \(x\in\left\{1;2\right\}\)
Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(A=\left|x-1\right|+\left|x-2\right|=\left|x-1\right|+\left|2-x\right|\ge\left|x-1+2-x\right|=\left|1\right|=1\)
Dấu " = " xảy ra khi \(\left\{{}\begin{matrix}x-1\ge0\\2-x\le0\end{matrix}\right.\Rightarrow1\le x\le2\)
Vì \(x\in Z\)
\(\Rightarrow x\in\left\{1;2\right\}\)
Vậy \(MIN_A=1\) khi x = 1 hoặc x = 2
