ta có:
A=|x-1|+|x-2|+|x-3|+|x-6|
=(|x-1|+|x-6|)+(|x-2|+|x-3|)
=(|x-1|+|6-x|)+(|x-2|+|3-x|)
≥ |x-1+6-x|+|x-2+3-x|
=5+1=6
=> Min A=6⇔$\left \{ {{(x-1)(6-x) ≥0} \atop {(x-2)(3-x)≥0}} \right.$ ⇔$\left \{ {{1≤x≤6} \atop {2≤x≤3}} \right.$ =>2≤x≤3
vậy Min A=6⇔2≤x≤3