\(B=\left|3x-1\right|+\left|4y+2\right|+3x\)(sửa đề)
\(B=\left|1-3x\right|+3x+\left|4y+2\right|\)
\(B\ge\left|1-3x+3x\right|+\left|4y+2\right|\)
\(B\ge1+\left|4y+2\right|\)
Vì \(\left|4y+2\right|\ge0\) nên \(B\ge1\)
Dấu "=" xảy ra khi:
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}1-3x\ge0\\3x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}1-3x\le0\\3x\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3x\le1\\3x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}3x\ge1\\3x\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow0\le3x\le1\Leftrightarrow0\le x\le\dfrac{1}{3}\)
Và \(\left|4y+2\right|=0\Leftrightarrow y=-\dfrac{1}{2}\)
