\(B=x-2\sqrt{x}+y-6\sqrt{y}+12\)
\(=x-2\sqrt{x}+1+y-6\sqrt{y}+9+2\)
\(=\left(\sqrt{x}-1\right)^2+\left(\sqrt{y}-3\right)^2+2\ge2\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\sqrt{x}-1=0\\\sqrt{y}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=9\end{matrix}\right.\)
Vậy \(Min_B=2\) khi x=1 và y=9