Question

tìm GTLN -8x^2+4xy-y^2+3

 

Answer 1

- Đặt \(A=-8x^2+4xy-y^2+3\)

\(\Rightarrow-A=8x^2-4xy+y^2-3\)

\(=\left(4x^2-4xy+y^2\right)+4x^2-3\)

\(=\left(2x-y\right)^2+4x^2-3\)

- Vì \(\left\{{}\begin{matrix}\left(2x-y\right)^2\ge0\\4x^2\ge0\end{matrix}\right.\)\(\Rightarrow\)\(-A=\left(2x-y\right)^2+4x^2-3\ge-3\)

\(\Rightarrow A\le3\)

- Vậy \(GTLN\) của A là \(3\), đạt tại \(\left\{{}\begin{matrix}\left(2x-y\right)^2=0\\4x^2=0\end{matrix}\right.\Leftrightarrow x=y=0\)