Sửa đề: \(...B=\left|x-2\right|+\left|y-3\right|\)
Giải:
Dễ thấy: \(\left\{{}\begin{matrix}\left|x-2\right|\ge0\forall x\\\left|y-3\right|\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left|x-2\right|+\left|y-3\right|\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left|x-2\right|=0\\\left|y-3\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y-3=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
Vậy \(B_{min}=0\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
ta có: \(\left|x-3\right|=\left|3-x\right|\)
\(B=\left|x-2\right|+\left|3-x\right|\ge\left|x-2+3-x\right|=1\)
dấu bằng xảy ra khi \(\left\{{}\begin{matrix}x-2< 0\\3-x< 0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x< 2\\x>-3\end{matrix}\right.\)\(\Leftrightarrow-3< x< 2\)
hoặc\(\left\{{}\begin{matrix}x-2>0\\3-x>0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x>2\\x< 3\end{matrix}\right.\)(loại)
vậy GTNN của B là 1 tại \(-3< x< 2\)
