\(A=x^2+2y^2-2xy-3y+2x-5=\left(x^2-2xy+y^2+2x-2y+1\right)+\left(y^2-y+\dfrac{1}{4}\right)+\dfrac{25}{4}=\left(x-y+1\right)^2+\left(y-\dfrac{1}{2}\right)^2+\dfrac{25}{4}\ge\dfrac{25}{4}\forall x;y\)
=>GTNN của A=\(\dfrac{25}{4}\)xảy ra khi \(\left\{{}\begin{matrix}x-y+1=0\\y-\dfrac{1}{2}=0\end{matrix}\right.\)<=>\(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{2}\end{matrix}\right.\)
\(A=x^2+2y^2-2xy-3y+2x-5\)
\(=\left(x^2+y^2+1-2xy-2y+2x\right)+y^2-\dfrac{1}{2}y.2+\dfrac{1}{4}-6,25\)
\(=\left(x-y+1\right)^2+\left(y-\dfrac{1}{2}\right)^2-6,25\ge-6,25\)
Dấu " = " khi \(\left\{{}\begin{matrix}\left(x-y+1\right)^2=0\\\left(y-\dfrac{1}{2}\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-1}{2}\\y=\dfrac{1}{2}\end{matrix}\right.\)
Vậy \(MIN_A=-6,25\) khi \(x=\dfrac{-1}{2}\) và \(y=\dfrac{1}{2}\)
\(A=x^2+y^2-2xy-3y+2x-5\)
\(=\left(x^2-2xy+2x+y^2-2y+1\right)+\left(y^2-y-6\right)\)
\(=\left(x-y-1\right)^2+\left(y^2-y-\dfrac{24}{4}\right)\)
\(=\left(x-y+1\right)^2+\left(y^2-y+\dfrac{1}{4}\right)-\dfrac{25}{4}\)
\(=\left(x-y+1\right)^2+\left(y-\dfrac{1}{2}\right)^2-\dfrac{25}{4}\ge\dfrac{-25}{4}\forall x,y\)