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