Ta có:\(B=\left|x-1\right|+\left|x-2\right|=\left|1-x\right|+\left|x-2\right|\)
\(\Rightarrow\left|1-x\right|+\left|x-2\right|\ge\left|1-x+x-2\right|\)\(=1\)
\(\Rightarrow\left|1-x\right|+\left|x-2\right|\ge1\)
Dấu" = "xảy ra khi:
TH1:\(\left\{{}\begin{matrix}1-x>0\\X-2>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< 1\\x>2\end{matrix}\right.\)ko thỏa mãn
TH2:\(\left\{{}\begin{matrix}1-x< 0\\x-2< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>1\\x< 2\end{matrix}\right.\Rightarrow1< x< 2\)
Vậy Bnn = 1 khi \(1< x< 2\)
\(\left|x-1\right|+\left|x-2\right|\)
=\(\left|1-x\right|+\left|x-2\right|\ge\left|1-x+x-2\right|=\left|-1\right|=1\)
\(\Rightarrow\left|x-1\right|+\left|x-2\right|\ge1\)
Dấu "=" xảy ra khi \(\left(1-x\right)\left(x-2\right)\ge0\)
\(\left\{{}\begin{matrix}\left\{{}\begin{matrix}1-x\ge0\\x-2\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}1-x\le0\\x-2\le0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x\le1\\x\ge2\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge1\\x\le2\end{matrix}\right.\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x\le1\\x\ge2\end{matrix}\right.\)(không có số nào thảo mãn)
\(\Leftrightarrow1\le x\le2\)
Vậy GTNN của\(\left|x-1\right|+\left|x-2\right|\) là 1 khi \(1\le x\le2\)