Question

Tìm a;b;c thỏa mãn đẳng thức

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

Answer 1

a² -2a+b²+4b+4c²-4c+6=0

<=>(a²-2a+1)+(b²+4b+4)+(4c²-4c+1)=0

<=>(a-1)²+(b+2)²+(2c-1)²=0

\(\left[{}\begin{matrix}a-1=0\\b+2=0\\2c-1=0\end{matrix}\right.\left[{}\begin{matrix}a=1\\b=-2\\c=\dfrac{1}{2}\end{matrix}\right.\)

Answer 2

Ta có: \(a^2-2a+b^2+4b+4c^2-4c+6=\)

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

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

Mà \(\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2\ge0\left(\forall a;b;c\right)\)

Dấu "=" xảy ra \(\Leftrightarrow\left(a-1\right)^2=0;\left(b+2\right)^2=0;\left(2c-1\right)^2=0\)

\(\Leftrightarrow a=1;b=-2;c=\dfrac{1}{2}\)