Đáp án: $A_{min}=2013⇔-2012≤x≤1$
Giải thích các bước giải:
Ta có: $A=|x-1|+|x+2012|$
$=|1-x|+|x+2012|≥|1-x+x+2012|=|2013|=2013$
Dấu bằng xảy ra $⇔(1-x)(x+2012)≥0$
$⇔\left[ \begin{array}{l}\large \left \{ {{1-x≥0} \atop {x+2012≥0}} \right.\\\large \left \{ {{1-x≤0} \atop {x+2012≤0}} \right.\end{array} \right.$
$⇔\left[ \begin{array}{l}\large \left \{ {{x≤1} \atop {x≥-2012}} \right.\\\large \left \{ {{x≥1} \atop {x≤-2012}} \right.\end{array} \right.$
$⇔-2012≤x≤1$
