a.
\(\Leftrightarrow\left(x^2+y^2+9+2xy-6x-6y\right)+3\left(9x^2+6x+1\right)\le3\)
\(\Leftrightarrow\left(x+y-3\right)^2+3\left(3x+1\right)^2\le3\)
\(\Rightarrow3\left(3x+1\right)^2\le3\)
\(\Rightarrow\left(3x+1\right)^2\le1\)
\(\Rightarrow\left[{}\begin{matrix}\left(3x+1\right)^2=0\\\left(3x+1\right)^2=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\left(ktm\right)\\x=0\\x=-\dfrac{2}{3}\left(ktm\right)\end{matrix}\right.\)
Thay \(x=0\Rightarrow y^2-6y+9\le0\)
\(\Rightarrow\left(y-3\right)^2\le0\Rightarrow\left(y-3\right)^2=0\)
\(\Rightarrow y=3\)
b.
\(\Leftrightarrow\left(4x^2+9y^2+25+12xy-20x-30y\right)+4x^2-4x+1\le4\)
\(\Leftrightarrow\left(2x+3y-5\right)^2+\left(2x-1\right)^2\le4\)
\(\Rightarrow\left(2x-1\right)^2\le4\)
\(\Rightarrow\left(2x-1\right)^2=1\) (do 2x-1 luôn lẻ)
\(\Rightarrow\left[{}\begin{matrix}2x-1=1\\2x-1=-1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\\end{matrix}\right.\)
- Với \(x=0\Rightarrow9y^2-30y+22\le0\)
\(\Rightarrow\left(3y-5\right)^2\le3\)
\(\Rightarrow\left[{}\begin{matrix}\left(3y-5\right)^2=0\\\left(3y-5\right)^2=1\end{matrix}\right.\) \(\Rightarrow y=...\)
- Với \(x=1\Rightarrow9y^2-18y+6\le0\)
\(\Rightarrow\left(3y-3\right)^2\le3\)
\(\Rightarrow\left(3y-3\right)^2=0\) (do 3y-3 luôn chia hết cho 3)
\(\Rightarrow y=1\)
