Đáp án:
Ta có: $y=|2a-1|+|a|≥|2a-1+a|=|3a-1|=x$
Dấu bằng xảy ra
$⇔(2a-1)a≥0$
$⇔\left[ \begin{array}{l}\left \{ {{2a-1≥0} \atop {a≥0}} \right.\\\left \{ {{2a-1≥0} \atop {a≥0}} \right.\end{array} \right.$
$⇔\left[ \begin{array}{l}\left \{ {{a≥0,5} \atop {a≥0}} \right.\\\left \{ {{a≤0,5} \atop {a≤0}} \right.\end{array} \right.$
$⇔\left[ \begin{array}{l}a≥0,5\\a≤0\end{array} \right.$