\(D=x^2+2y^2-2xy+4x-3y-12\)
\(=\left(x^2-2xy+4x\right)+2y^2-3y-12\)
\(=\left[x^2-2x\left(y-2\right)+\left(y-2\right)^2\right]+2y^2-3y-12-y^2+4y-4\)\(=\left(x-y+2\right)^2+y^2+y+8\)
\(=\left(x-y+2\right)^2+\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{31}{4}\)
\(=\left(x-y+2\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{31}{4}\ge\dfrac{31}{4}\forall x\)
Vậy Min D = \(\dfrac{31}{4}\) khi \(\left\{{}\begin{matrix}x-y+2=0\\y+\dfrac{1}{2}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+\dfrac{5}{2}=0\\y=-\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\)
Đúng 0
Bình luận (0)
