$H=|x-1|+|x-4|=|x-1|+|4-x|≥|x-1+4-x|=3∀x$
Dấu "=" xảy ra
\(⇔(x-1)(4-x)≥0⇔\left[ \begin{array}{l}\left \{ {{x-1≥0} \atop {4-x≥0}} \right.\\\left \{ {{x-1≤0} \atop {4-x≤0}} \right.\end{array} \right.⇔\)\(\left[ \begin{array}{l}\left \{ {{x≥1} \atop {x≤4}} \right.\\\left \{ {{x≤1} \atop {x≥4}} \right.\end{array} \right.⇔\) \(\left[ \begin{array}{l}1≤x≤4\\x∈∅\end{array} \right.\)
Vậy .....
$K=|x-50|+|x-30|=|50-x|+|x-30|≥|50-x+x-30|=20∀x$
(Rồi bn lý luận như trên nha)
