\(=\left(x^2-4xy+4y^2\right)+10\left(x-2y\right)+25+\left(y^2-2y+1\right)+2\\ =\left(x-2y\right)^2+10\left(x-2y\right)+25+\left(y-1\right)^2+2\\ =\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-2y+5=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+3=0\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)

Vậy GTNN của biểu thức là 2

\(C=x^2-4xy+5y^2+10x-22y+28\)

    \(=x^2-4xy+10x+4y^2+25-10y+y^2-2y+3\)

    \(=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)

Vậy \(GTNN=2\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)

\(G=x^2-4xy+5y^2+10x-22y+28.\)

\(=\left(x^2+4y^2+25-4xy+10x-20y\right)+\left(y^2-2y+1\right)+2\)

\(=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\)

Do \(\left(x-2y+5\right)^2+\left(y-1\right)^2\ge0\forall x\)nên \(\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)

Vậy \(MinG=2\Leftrightarrow\hept{\begin{cases}\left(x-2y+5\right)^2=0\\\left(y-1\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}}\)

\(G=x^2-4xy+5y^2+10x-22y+28\)

\(G=x^2-2x\left(2y-5\right)+5y^2-22y+28\)

\(G=x^2-2x\left(2y-5\right)+\left(4y^2-20y+25\right)+\left(y^2-2y+1\right)+2\)

\(G=x^2-2x\left(2y-5\right)+\left(2x-5\right)^2+\left(y-1\right)^2+2\)

\(G=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)

Dấu "=" xảy ra khi x=-3;y=1