Lời giải:
\(A=3x^2+11y^2-2xy-2x+6y-1\)
\(\Leftrightarrow A=\left(x^2+y^2+\frac{1}{4}-2xy-x+y\right)+2\left(x^2-\frac{1}{2}x+\frac{1}{16}\right)+10\left(y^2+\frac{1}{2}y+\frac{1}{16}\right)-2\)
\(\Leftrightarrow A=\left(x-y-\frac{1}{2}\right)^2+2\left(x-\frac{1}{4}\right)^2+10\left(y+\frac{1}{4}\right)^2-2\)
Thấy rằng \(\hept{\begin{cases}\left(x-y-\frac{1}{2}\right)^2\ge0\\\left(x-\frac{1}{4}\right)^2\ge0\\\left(y+\frac{1}{4}\right)^2\ge0\end{cases}}\Rightarrow A\ge-2\)
Vậy \(A_{min}=-2\Leftrightarrow\hept{\begin{cases}x-y-\frac{1}{2}=0\\x-\frac{1}{4}=0\\y+\frac{1}{4}=0\end{cases}\Leftrightarrow x=\frac{1}{4};y=\frac{-1}{4}}\)
