$A={{x}^{2}}+2x+{{y}^{2}}-4y+9$$=\left( {{x}^{2}}+2x+1 \right)+\left( {{y}^{2}}-4y+4 \right)+4$
$={{\left( x+1 \right)}^{2}}+{{\left( y-2 \right)}^{2}}+4$
Ta có: $\left\{ \begin{align} & {{\left( x+1 \right)}^{2}}\ge 0 \\ & {{\left( y-2 \right)}^{2}}\ge 0 \\ \end{align} \right.,\forall x,y$$\Rightarrow {{\left( x+1 \right)}^{2}}+{{\left( y-2 \right)}^{2}}+4\ge 4,\forall x,y$
$\Rightarrow $GTNN của A là 4. Dấu “=” xảy ra $\Leftrightarrow \left\{ \begin{align} & {{\left( x+1 \right)}^{2}}=0 \\ & {{\left( y-2 \right)}^{2}}=0 \\ \end{align} \right.$$\Leftrightarrow \left\{ \begin{align} & x=-1 \\ & y=2 \\ \end{align} \right.$