\(=\left(4x^2-4xy+y^2\right)+\left(x^2+2x+1\right)+\left(3y^2-12y+12\right)-7\)
\(=\left(2x-y\right)^2+\left(x+1\right)^2+\left(\sqrt{3}y-\dfrac{12}{2\sqrt{3}}\right)^2-7\ge-7\)
\(\Rightarrow\) \(minD=-7\) khi \(\left\{{}\begin{matrix}\left(2x-y\right)^2=0\\\left(x+1\right)^2=0\\\left(\sqrt{3}y-\dfrac{12}{2\sqrt{3}}\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}2x-y=0\\x+1=0\\\sqrt{3}y-\dfrac{12}{2\sqrt{3}}=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-1\\2\left(-1\right)-y=0\\\sqrt{3}y=\dfrac{12}{2\sqrt{3}}\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-1\\y=-2\\y=2\end{matrix}\right.\) \(\Rightarrow\) đề sai
D = 5x2 - 4xy + 4y2 + 2x - 12y + 6
= (x2 + 4y2 - 4xy) + 9 + (6x - 12y) + (4x2 - 4x + 1) - 4
= (x - 2y)2 + 32 + 2.(x - 2y).3 + (2x - 1)2 - 4
= (x - 2y + 3)2 + (2x - 1)2 - 4
Mà \(\left(x-2y+3\right)^2\ge0\forall x;y\); \(\left(2x-1\right)^2\ge0\forall x\) nên
D \(\ge-4\forall x;y\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x-2y+3=0\\2x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{7}{4}\\x=\dfrac{1}{2}\end{matrix}\right.\)
