Ta có:$P=x^2-4x+4y^2+12y+13$ $=(x-2)^2+(2y+3)^2\ge0$$\min P=0 \Leftrightarrow \left\{\begin{array}{l}x=2\\y=\dfrac{-2}{3}\end{array}\right.$
Ta có:$P=x^2-4x+4y^2+12y+13$ $=(x-2)^2+(2y+3)^2\ge0$$\min P=0 \Leftrightarrow \left\{\begin{array}{l}x=2\\y=\dfrac{2}{3}\end{array}\right.$
Ta có:$P=x^2-4x+4y^2+12y+13$ $=(x-2)^2+(2y+3)^2\ge0$$\min P=0 \Leftrightarrow \left\{\begin{array}{l}x=2\\y=\dfrac{
-2}{3}\end{array}\right.$