Ta thấy: \(\left\{{}\begin{matrix}\sqrt{x-2y+1}\ge0\\\left(x-3y\right)^{2012}\ge0\end{matrix}\right.\)\(\forall x,y\)
\(\Rightarrow\sqrt{x-2y+1}+\left(x-3y\right)^{2012}\ge0\forall x,y\)
\(\Rightarrow\sqrt{x-2y+1}+\left(x-3y\right)^{2012}+3\ge3\forall x,y\)
\(\Rightarrow B\ge3\forall x,y\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}\sqrt{x-2y+1}=0\\\left(x-3y\right)^{2012}=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x-2y+1=0\\x-3y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2y-1\\x=3y\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-1\end{matrix}\right.\)
Vậy với \(\left\{{}\begin{matrix}x=-3\\y=-1\end{matrix}\right.\) thì \(B_{Min}=3\)
ta có: \(\sqrt{\left(x-2y+1\right)}\ge0\)
\(\left(x-3y\right)\ge0\)
B min =0+0+3=3
