Câu 2:

\(a^2-2a+b^2+4b+4c^2-4c+6=0\\ \Leftrightarrow\left(a^2-2a+1\right)+\left(b^2+4b+4\right)+\left(4c^2-4c+1\right)=0\\ \Leftrightarrow\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2=0\\ Do\text{ }\left(a-1\right)^2\ge0\forall x\\ \left(b+2\right)^2\ge0\forall x\\ \left(2c-1\right)^2\ge0\forall x\\ \Leftrightarrow\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2\ge0\forall x\\ \text{Dấu }"="\text{ xảy ra khi }:\left\{{}\begin{matrix}\left(a-1\right)^2=0\\\left(b+2\right)^2=0\\\left(2c-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a-1=0\\b+2=0\\2c-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=-2\\c=\dfrac{1}{2}\end{matrix}\right.\)

Vậy \(a=1;b=-2;c=\dfrac{1}{2}\)