\(a^2+b^2+4c^2=2a-4b+4c-6\)
\(\Leftrightarrow a^2+2a+1+b^2+4b+4+4c^2-4c+1=0\)
\(\Leftrightarrow\left(a+1\right)^2+\left(b+2\right)^2+4\left(c^2-2.c.\dfrac{1}{2}+\dfrac{1}{4}\right)=0\)
\(\Leftrightarrow\left(a+1\right)^2+\left(b+2\right)^2+4\left(c-\dfrac{1}{2}\right)^2=0\)
Mà \(\left\{{}\begin{matrix}\left(a+1\right)^2\ge0\\\left(b+2\right)^2\ge0\\4\left(c-\dfrac{1}{2}\right)^2\ge0\end{matrix}\right.\Rightarrow\left(a+1\right)^2+\left(b+2\right)^2+4\left(c-\dfrac{1}{2}\right)^2\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(a+1\right)^2=0\\\left(b+2\right)^2=0\\4\left(c-\dfrac{1}{2}\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=-1\\b=-2\\c=\dfrac{1}{2}\end{matrix}\right.\)
Vậy \(a=-1,b=-2,c=\dfrac{1}{2}\)
