\(P=x(x-3)+y(y-3)+xy+2014\)
\(P=x^2-3x+y^2-3y+xy+2014\)
\(\Rightarrow4P=4x^2-12x+4y^2-12y+4xy+8056\)
\(=\left(4x^2+4xy+y^2\right)-12x-6y+3y^2-6y+8056\)
\(=\left(2x+y\right)^2-6\left(2x+y\right)+9+3\left(y^2-2y+1\right)+8044\)
\(=\left(2x+y-3\right)^2+3\left(y-1\right)^2+8044\ge8044\)
\(\Rightarrow P\ge\frac{8044}{2}=2011\)
Dấu \("="\) xảy ra \(\left\{{}\begin{matrix}\left(y-1\right)^2=0\\\left(2x+y-3\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=1\end{matrix}\right.\)
Đúng 0
Bình luận (0)