\(\left(x+\dfrac{1}{2}\right)^2+\left(y-3\right)=0\)
Để GTBT = 0 \(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{2}=0\\y-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=3\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(-\dfrac{1}{2};3\right)\) thì GTBT = 0.