Ta có:
\(\hept{\begin{cases}\left|3x-5\right|=\left|5-3x\right|\ge5-3x\\\left|2-3x\right|=\left|3x-2\right|\ge3x-2\end{cases}}\)
\(\Rightarrow\left|5-3x\right|+\left|3x-2\right|\ge\left(5-3x\right)+\left(3x-2\right)\)
\(\Rightarrow\left|3x-5\right|+\left|2-3x\right|\ge3\)
\(\Rightarrow B\ge3\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}5-3x\ge0\\3x-2\ge0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}3x\le5\\3x\ge2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\le\frac{5}{3}\\x\ge\frac{2}{3}\end{cases}}\)
\(\Leftrightarrow\frac{2}{3}\le x\le\frac{5}{3}\)
Vậy MinB = 3 \(\Leftrightarrow\frac{2}{3}\le x\le\frac{5}{3}\)
