Question

Tìm giá trị nhỏ nhất:

\(P=x^2-2xy+2y^2-2x+3y+3\)

Answer 1

\(P=x^2-2xy+2y^2-2x+3y+3\)

\(=x^2-2x\left(y+1\right)+\left(y+1\right)^2-\left(y+1\right)^2+2y^2+3y+3\)

\(=\left(x-y-1\right)^2+y^2+y+2\)

\(=\left(x-y-1\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{7}{4}\)

\(Vì\) \(\left(x-y-1\right)^2+\left(y+\dfrac{1}{2}\right)^2\ge0\forall x,y\)

\(MinP=\dfrac{7}{4}\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\)