Question

Tìm Min:

D=\(6x-18y-x^2-13y^2+20+4xy\)

Answer 1

- Sửa đề: Tìm max.

\(D=6x-18y-x^2-13y^2+20+4xy\)

\(=-\left(x^2+4y^2+9-6x+12y-4xy\right)-\left(9y^2+6y+1\right)+30\)

\(=-\left[\left(x-2y\right)^2-6\left(x-2y\right)+9\right]-\left(3y+1\right)^2+30\)

\(=-\left(x-2y-3\right)^2-\left(3y+1\right)^2+30\le30\)

- Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-2y-3=0\\3y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7}{3}\\y=-\dfrac{1}{3}\end{matrix}\right.\)

- Vậy \(MaxD=30\)