\(3x^2+3y^2+6x-12y+15=\left(3x^2+6x+3\right)+\left(3y^2-12y+12\right)\)
\(=3.\left(x^2+2x+1\right)+3.\left(y^2-4y+4\right)\)
\(=3.\left(x+1\right)^2+3.\left(y-2\right)^2\)
\(=3.\left(\left(x+1\right)^2+\left(y-2\right)^2\right)\)
\(\Rightarrow3.\left(\left(x+1\right)^2+\left(y-2\right)^2\right)=0\Rightarrow\left(x+1\right)^2+\left(y-2\right)^2=0\)
Mà \(\left(x+1\right)^2\ge0,\forall x\inℝ\)
\(\left(y-2\right)^2\ge0,\forall y\inℝ\)
\(\Rightarrow\left(x+1\right)^2+\left(y-2\right)^2\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x+1\right)^2=0\\\left(y-2\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+1=0\\y-2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-1\\y=2\end{cases}}}\)
