$A=x^2+9y^2-6xy+4x-12y+4+x^2-10x+25+1979$$=(x-3y+2)^2+(x-2)^2+1979 \geq 1979$Dấu $"="$ xảy ra khi $\begin{cases}x-3y+2=0 \\ x-2=0 \end{cases}\Leftrightarrow \begin{cases}x=2 \\ y= \frac{4}{3}\end{cases}$
$A=x^2+9y^2-6xy+4x-12y+4+x^2-10x+25+197
5$$=(x-3y+2)^2+(x-
5)^2+1979 \geq 1979$Dấu $"="$ xảy ra khi $\begin{cases}x-3y+2=0 \\ x-
5=0 \end{cases}\Leftrightarrow \begin{cases}x=
5 \\ y= \frac{
7}{3}\end{cases}$