Giải

a)$A={{x}^{2}}-4xy+5{{y}^{2}}+10x-22y+28$

$A=({{x}^{2}}+4{{y}^{2}}+25-4xy+10x-20y)+({{y}^{2}}-2y+1)+2$

$A=[{{x}^{2}}+{{(-2y)}^{2}}+{{5}^{2}}+2.x.(-2y)+2.x.5+2.(-2y).5]+({{y}^{2}}-2y+1)+2$

$A={{(x-2y+5)}^{2}}+{{(y-1)}^{2}}+2$

Ta có :${{(x-2y+5)}^{2}}\ge 0\,;{{(y-1)}^{2}}\ge 0\,\forall x,y$

Suy ra $A={{(x-2y+5)}^{2}}+{{(y-1)}^{2}}+2\ge 2$

Dấu “ = ” xảy ra $\Leftrightarrow \left\{ \begin{align}& {{(x-2y+5)}^{2}}=0 \\ & {{(y-1)}^{2}}=0 \\ \end{align} \right.$$\Leftrightarrow \left\{ \begin{align}& x=-3 \\ & y=1 \\ \end{align} \right.$

Vậy GTNN của A = 2 khi $(x;y)=(-3;1)$

b) $B=-{{x}^{2}}-2{{y}^{2}}+2xy-4x+6y-9$

$B=-({{x}^{2}}+2{{y}^{2}}-2xy+4x-6y+9)$

$B=-[({{x}^{2}}+{{y}^{2}}+{{2}^{2}}-2.x.y+2.x.2-2.y.2)+({{y}^{2}}-2y+1)]+4$

$B=-[{{(x-y+2)}^{2}}+{{(y-1)}^{2}}+4]$

$-B={{(x-y+2)}^{2}}+{{(y-1)}^{2}}+4$

Ta có :${{(x-y+2)}^{2}}\ge 0\,;{{(y-1)}^{2}}\ge 0\,\forall x,y$

Suy ra : $-B={{(x-y+2)}^{2}}+{{(y-1)}^{2}}+4\ge 4$

Do đó : $B=-[{{(x-y+2)}^{2}}+{{(y-1)}^{2}}+4]\le -4$

Dấu “ = ” xảy ra $\Leftrightarrow \left\{ \begin{align}& {{(x-y+2)}^{2}}=0 \\ & {{(y-1)}^{2}}=0 \\ \end{align} \right.$$\Leftrightarrow \left\{ \begin{align} & x=-1 \\
& y=1 \\ \end{align} \right.$

Vậy GTLN của B = $-4$ khi $(x;y)=(-1;1)$

c)$C={{x}^{2}}+{{y}^{2}}+xy+x+y$

$4C=4{{x}^{2}}+4{{y}^{2}}+4xy+4x+4y$

$4C=(4{{x}^{2}}+{{y}^{2}}+1+4xy+4x+2y)+(3{{y}^{2}}+2y-1)$

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

Ta có :${{(x+y+1)}^{2}}\ge 0\,;{{\left( y+\frac{1}{3} \right)}^{2}}\ge 0\,\forall x,y$

Suy ra $4C={{(2x+y+1)}^{2}}+3{{\left( y+\frac{1}{3} \right)}^{2}}-\frac{4}{3}\ge -\frac{4}{3}$

Do đó $C\ge -\frac{1}{3}$

Dấu “ = ” xảy ra $\Leftrightarrow \left\{ \begin{align}& {{(2x+y+1)}^{2}}=0 \\ & {{\left( y+\frac{1}{3} \right)}^{2}}=0 \\ \end{align} \right.$$\Leftrightarrow \left\{ \begin{align}& x=-\frac{1}{3} \\ & y=-\frac{1}{3} \\
\end{align} \right.$

Vậy GTNN của C = $-\frac{1}{3}$ khi $(x;y)=\left( -\frac{1}{3};-\frac{1}{3} \right)$