a)Ta thấy:
\(\left|x+1,5\right|\ge0\)
\(\Rightarrow A\ge0\)
Dấu = khi x=-1,5
Vậy MinA=0 khi x=-1,5
b)Ta thấy:\(\left|x-\frac{1}{2}\right|\ge0\)
\(\Rightarrow\left|x-\frac{1}{2}\right|-\frac{9}{10}\ge-\frac{9}{10}\)
\(\Rightarrow B\ge-\frac{9}{10}\)
Dấu = khi \(x=\frac{1}{2}\)
Vậy MinB=\(-\frac{9}{10}\) khi \(x=\frac{1}{2}\)