\(S_n=\sqrt{\left(x-n\right)^2}=\left|x-n\right|\)
\(f\left(x\right)=\left|x-1\right|+\left|x-2\right|+\left|x-3\right|=\left|x-1\right|+\left|3-x\right|+\left|x-2\right|\)
\(\ge\left|x-1+3-x\right|+0=2\)
Dấu "=" xảy ra khi \(\left(x-1\right)\left(3-x\right)\ge0\text{ và }x-2=0\Leftrightarrow x=2\)