\(5x^2+5y^2-5x-15y+8\le0\)
\(\Leftrightarrow10x^2+10y^2-10\left(x+3y\right)+16\le0\)
\(\Rightarrow\left(9x^2-6xy+y^2\right)+\left(x^2+6xy+9y^2\right)-10\left(x+3y\right)+16\le0\)
\(\Leftrightarrow\left(3x-y\right)^2+\left(x+3y\right)^2-10\left(x+3y\right)+16\le0\)
Do \(\left(3x-y\right)^2\ge0;\forall x;y\)
\(\Rightarrow\left(x+3y\right)^2-10\left(x+3y\right)+16\le0\)
\(\Leftrightarrow\left(x+3y-5\right)^2\le9\)
\(\Rightarrow-3\le x+3y-5\le3\)
\(\Rightarrow2\le x+3y\le8\)
Dấu "=" (bên phải) xảy ra khi \(\left\{{}\begin{matrix}3x-y=0\\x+3y=8\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{4}{5}\\y=\dfrac{12}{5}\end{matrix}\right.\)