dùng HĐT

\(=\left(4x^2-4xy+y^2\right)+\left(x^2+2x+1\right)+\left(3y^2-12y+12\right)-7\)

\(=\left(2x-y\right)^2+\left(x+1\right)^2+\left(\sqrt{3}y-\dfrac{12}{2\sqrt{3}}\right)^2-7\ge-7\)

\(\Rightarrow\) \(minD=-7\) 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 \(\left\{{}\begin{matrix}x-2y+3=0\\2x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{7}{4}\\x=\dfrac{1}{2}\end{matrix}\right.\)

\(A=2x^2+4y^2+4xy+10x+12y+18\)

\(A=x^2+4xy+4y^2+6x+12y+9+x^2+4x+4+5\)

\(A=\left(x+2y\right)^2+2.3\left(x+2y\right)+9+\left(x+2\right)^2+5\)

\(A=\left(x+2y+3\right)^2+\left(x+2\right)^2+5\)

Do : \(\hept{\begin{cases}\left(x+2y+3\right)^2\ge0\forall x\\\left(x+2\right)^2\ge0\forall x\end{cases}}\)

\(\Leftrightarrow\left(x+2y+3\right)^2+\left(x+2\right)^2+5\ge5\)

\("="\Leftrightarrow\hept{\begin{cases}x+2y+3=0\\x+2=0\end{cases}\Leftrightarrow\hept{\begin{cases}y=-\frac{1}{2}\\x=-2\end{cases}}}\)

Vậy \(A_{min}=5\Leftrightarrow\hept{\begin{cases}x=-2\\y=-\frac{1}{2}\end{cases}}\)

Chúc bạn học tốt !!!