Ta có:\(\left\{{}\begin{matrix}-\left(x-\sqrt{5}\right)^2\le0\forall x\\-\left(y+\sqrt{2}\right)^2\le0\forall y\end{matrix}\right.\)
=> \(-\sqrt{7}-\left(x-\sqrt{5}\right)^2-\left(y+\sqrt{2}\right)^2\le-\sqrt{7}\)
Dấu ''='' xảy ra khi: \(\left\{{}\begin{matrix}x-\sqrt{5}=0\\y+\sqrt{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{5}\\y=-\sqrt{2}\end{matrix}\right.\)
Vậy GTLN của B = \(-\sqrt{7}\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{5}\\y=-\sqrt{2}\end{matrix}\right.\)