\(N=2013-\left(x^2+2xy+y^2\right)-\left(y^2-6x+9\right)\)
\(N=2013-\left(x+y\right)^2-\left(y-3\right)^2\le2013-0-0=2013\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}y-3=0\\x+y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}y=3\\x+3=0\end{cases}}\Leftrightarrow x=-3;y=3\)
khó thế :Đ