Question

Tìm x để y đạt GTLN thỏa mãn \(x^2+2y^2+2xy-8x-4y=0\)

Answer 1

\(x^2+2\left(y-4\right)x+2y^2-4y=0\)

\(\Delta'=\left(y-4\right)^2-\left(2y^2-4y\right)\ge0\)

\(\Leftrightarrow-y^2-4y+16\ge0\)

\(\Rightarrow-2-2\sqrt{5}\le y\le-2+2\sqrt{5}\)

\(\Rightarrow y_{max}=-2+2\sqrt{5}\)

Khi đó \(x=6-2\sqrt{5}\)

Answer 2

Ta có :

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

\(=\left(x^2+2xy+y^2\right)-8\left(x+y\right)+16+\left(y^2-4y+4\right)-20\)

\(=\left(x+y-4\right)^2+\left(y-2\right)^2-20\)

Với mọi x ta có :

\(\left\{{}\begin{matrix}\left(x+y-4\right)^2\ge0\\\left(y-2\right)^2\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left(x+y-4\right)^2+\left(y-2\right)^2\ge0\)

\(\Leftrightarrow A\ge-20\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=2\)

Vậy..