\(a^2-2a+b^2+4b+4c^2-4c+6\)
\(=\left(a^2-2a+1\right)+\left(b^2+4b+4\right)+\left(4c^2-4c+1\right)\)
\(=\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2\)
\(\Rightarrow\begin{cases}\left(a-1\right)^2\ge0\\\left(b+2\right)^2\ge0\\\left(2c-1\right)^2\ge0\end{cases}\)
\(\Rightarrow\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2\ge0\)
Dấu = khi \(\begin{cases}\left(a-1\right)^2=0\\\left(b+2\right)^2=0\\\left(2c-1\right)^2=0\end{cases}\)\(\Leftrightarrow\begin{cases}a=1\\b=-2\\c=\frac{1}{2}\end{cases}\)
Vậy \(\begin{cases}a=1\\b=-2\\c=\frac{1}{2}\end{cases}\)
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