a,Đặt A= \(2x^2-4xy+4y^2-6x\)

\(=\left(2x^2-4xy-6x\right)+4y^2\)

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

\(=2\left[x^2-2x\left(y+\dfrac{3}{2}\right)+\left(y+\dfrac{3}{2}\right)^2\right]+4y^2-\left(y+\dfrac{3}{2}\right)^2\)

\(=2\left(x-y-\dfrac{3}{2}\right)^2+4y^2-y^2-3y-\dfrac{9}{4}\)

\(=2\left(x-y-\dfrac{3}{2}\right)^2+3\left(y^2-y+\dfrac{1}{4}\right)-3\)

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

Với mọi giá trị của x;y ta có:

\(\left(x-y-\dfrac{3}{2}\right)^2\ge0;\left(y-\dfrac{1}{2}\right)^2\ge0\)

\(\Rightarrow2\left(x-y-\dfrac{3}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2-3\ge-3\)

Vậy Min A = -3 khi \(\left\{{}\begin{matrix}x-y-\dfrac{3}{2}=0\\y-\dfrac{1}{2}=0\end{matrix}\right.\)

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

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

b, Đặt B = \(z^2-4zt+5t^2-2t+13\)

\(=\left(z^2-4zt+4t^2\right)+\left(t^2-2t+1\right)+12\)

\(=\left(z-2t\right)^2+\left(t-1\right)^2+12\)

Với mọi giá trị của z;t ta có:

\(\left(z-2t\right)^2\ge0;\left(t-1\right)^2\ge0\)

\(\Rightarrow\left(z-2t\right)^2+\left(t-1\right)^2+12\ge12\)

Vậy Min B = 12 khi \(\left\{{}\begin{matrix}z-2t=0\\t-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}z-2=0\\t=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}z=2\\t=1\end{matrix}\right.\)

c, Đặt C = \(16x^2-8x+y^2-2y\)

\(=\left(16x^2-8x+1\right)+\left(y^2-2y+1\right)-2\)

\(=\left(4x-1\right)^2+\left(y-1\right)^2-2\)

Với mọi giá trị x;y ta có:

\(\left(4x-1\right)^2\ge0;\left(y-1\right)^2\ge0\)

\(\Rightarrow\left(4x-1\right)^2+\left(y-1\right)^2-2\ge-2\)

Vậy Min C = -2 khi \(\left\{{}\begin{matrix}4x-1=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x=1\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4}\\y=1\end{matrix}\right.\)