Tìm a, b, c biết: a^2 - 2a + b^2 + 4b + 4c^2 - 4c + 6 = 0

* Hướng dẫn giải

Ta có:
$\begin{array}{l}{a}^2-2a+b^2+4b+4c^2-4c+6=0\\ \left(a^2-2a+1\right)+\left(b^2+4b+4\right)+\left(4c^2-4c+1\right)=0\\ {\left(a-1\right)}^2+{\left(b+2\right)}^2+{\left(2c-1\right)}^2=0\\ Do{\left(a-1\right)}^2\ge 0,{\left(b+2\right)}^2\ge 0;{\left(2c-1\right)}^2\ge 0\\ \Rightarrow \left\{\begin{array}{l}a-1=0\\ b+2=0\\ 2c-1=0\end{array}\right.\iff \left\{\begin{array}{l}a=1\\ b=-2\\ c=\frac{1}{2}\end{array}\right.\end{array}$

Vậy $a=1;b=-2;c=\frac{1}{2}$