Giải thích các bước giải:
$P=x^2+3y^2-2xy+2x-4y+5$
$\rightarrow P=(x^2-2x(y-1)+((y-1)^2)+2y^2-2y+\dfrac{1}{2})+\dfrac{1}{2}+3$
$\rightarrow P=(x-y+1)^2+2(y^2-2y.\dfrac{1}{2}+\dfrac{1}{4})+\dfrac{7}{2}$
$\rightarrow P=(x-y+1)^2+2(y-\dfrac{1}{2})^2+\dfrac{7}{2}$
$\rightarrow P\ge \dfrac{7}{2}\quad\forall x,y$
Dấu = xảy ra $\begin{cases}x-y+1=0\\ y-\dfrac{1}{2}=0\end{cases}\rightarrow \begin{cases}x=-\dfrac{1}{2}\\ y=\dfrac{1}{2}\end{cases}$
