\(A=\text{|}x-1\text{|}+\text{|}x-4\text{|}=\text{|}x-1\text{|}+\text{|}4-x\text{|}\) ≥ \(\text{|}x-1+4-x\text{|}=3\)
⇒ \(A_{MIN}=3."="\) xảy ra khi và chỉ khi : \(x-1\text{≥}0;4-x\text{≥}0\) ⇔ 1 ≤ x ≤ 4
|x-1+|x-4|=|x-1|+|4-x| ≥ |x-1+4-x|= |3|=3
Min =3 khi (x-1)(4-x) ≥ 0<=> 1≤ x≤ 4