Question

Tìm a,b,c để :

\(a^2-2ab+b^2+4b+4c^2-4c+6=0\)

 

Answer 1

Ta có :

\(a^2-2a+b^2+4b+4c^2-4c+6=0\)

\(\Leftrightarrow\left(x^2-2a+1\right)+\left(b^2+4b+4\right)+\left(4c^2-4c+1\right)=0\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b+2\right)^2+\left(2c+1\right)^2=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}a-1=0\\b+2=0\\2c+1=0\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}a=1\\b=-2\\c=-\frac{1}{2}\end{array}\right.\)

Vậy ..................