Ta có:
\(x^2+3y^2+2xy-10x-14y+18=0\)
\(\Leftrightarrow\left(x^2+2xy+y^2-10x-10y+25\right)+2y^2-4y-7=0\)
\(\Leftrightarrow\left(x+y\right)^2-2.\left(x+y\right).5+25+2y^2-4y-7=0\)
\(\Leftrightarrow\left(x+y-5\right)^2+2y^2-4y+2-9=0\)
\(\Leftrightarrow\left(x+y-5\right)^2+2\left(y^2-2y+1\right)=9\)
\(\Leftrightarrow\left(x+y-5\right)^2+2\left(y-1\right)^2=9\)
Vì \(2\left(y-1\right)^2\ge0\forall y\) nên \(\left(x+y-5\right)^2\le9\).
\(\Rightarrow-3\le x+y-5\le3\)
\(\Rightarrow2\le x+y\le8\)
Tới đây bạn suy ra GTNN và GTLN rồi tính 5x + 5y bình thường.