Với \(\forall x;y\) ta có :
\(\left\{{}\begin{matrix}\left(x+y-3\right)^4\ge0\\\left(x-2y\right)^2\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left(x+y-3\right)^4+\left(x-2y\right)^2\ge0\)
\(\Leftrightarrow\left(x+y-3\right)^4+\left(x-2y\right)^2+2012\ge2012\)
\(\Leftrightarrow Q\ge2012\)
Dấu "=" xảy ra khi :
\(\left\{{}\begin{matrix}\left(x+y-3\right)^4=0\\\left(x-2y\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y-3=0\\x-2y=0\end{matrix}\right.\)
Ta có :
\(\left(x+y-3\right)^4\ge0\\ \left(x-2y\right)^2\ge0\\ \Rightarrow Q\ge2012\\ \)
Dấu "=" xảy ra khi
\(\left\{{}\begin{matrix}x+y-3=0\\x-2y=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x+y=3\\x=2y\end{matrix}\right.\\ \Rightarrow3y=3\\ \Rightarrow y=1\\ \Rightarrow x=2\)
Min A = 2012 khi x=2;y=1
