a, \(A=\left|x-1\right|+\left|x-2\right|=\left|x-1\right|+\left|2-x\right|\ge\left|x-1+2-x\right|=1\)
Dấu "=" xảy ra khi \(\left(x-1\right)\left(2-x\right)\ge0\Leftrightarrow1\le x\le2\)
Vậy GTNN của A = 1 khi \(1\le x\le2\)
b, \(B=\left|x-1\right|+\left|x-2\right|+\left|x-3\right|=\left(\left|x-1\right|+\left|x-3\right|\right)+\left|x-2\right|\)
Ta có: \(\left|x-1\right|+\left|x-3\right|=\left|x-1\right|+\left|3-x\right|\ge\left|x-1+3-x\right|=2\)
Mà \(\left|x-2\right|\ge0\)
\(\Rightarrow B=\left|x-1\right|+\left|x-2\right|+\left|x-3\right|\ge2+0=2\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}\left(x-1\right)\left(3-x\right)\ge0\\\left|x-2\right|\ge0\end{cases}\Rightarrow x=2}\)
Vậy GTNN của B = 2 khi x = 2
c, \(C=\left|x-1\right|+\left|x-2\right|+\left|x-3\right|+\left|x-4\right|\)
\(=\left(\left|x-1\right|+\left|3-x\right|\right)+\left(\left|x-2\right|+\left|4-x\right|\right)\)
\(\ge\left|x-1+3-x\right|+\left|x-2+4-x\right|\)
\(\ge2+2=4\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}\left(x-1\right)\left(3-x\right)\ge0\\\left(x-2\right)\left(4-x\right)\ge0\end{cases}\Rightarrow\hept{\begin{cases}1\le x\le3\\2\le x\le4\end{cases}\Rightarrow}2\le x\le}3\)
Vậy GTNN của C = 4 khi \(2\le x\le3\)
