\(=\left(x^2-3x+\frac{9}{4}\right)+2\left(y^2-2y+1\right)+10-\frac{9}{4}-2\)
\(=\left(x-\frac{3}{2}\right)^2+2\left(y-1\right)^2+\frac{23}{4}\ge\frac{23}{4}\)
Dấu "=" xảy ra khi \(\begin{cases}\left(x-\frac{3}{2}\right)^2=0\\\left(y-1\right)^2=0\end{cases}\) \(\Leftrightarrow\begin{cases}x=\frac{3}{2}\\y=1\end{cases}\)
Vậy Min Q = \(\frac{23}{4}\) tại (x;y) = (\(\frac{3}{2};1\))
E đề ghi không rõ ...