\(\left(a^2-2a+1\right)+\left(b^2+4b+4\right)+\left(4c^2-4c+1\right)=0.\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b+2\right)^2+\left(2b-1\right)^2=0\)
Mà \(\left(a-1\right)^2\ge0\forall a\), \(\left(b+2\right)^2\ge0\forall b\),\(\left(2c-1\right)^2\ge0\forall c\)
\(\Rightarrow\hept{\begin{cases}\left(a-1\right)^2=0\\\left(b+2\right)^2=0\\\left(2c-1\right)^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}a=1\\b=-2\\c=\frac{1}{2}\end{cases}}.\)
a2 - 2a + b2 + 4b + 4c2 - 4c + 6 = 0
\(\Leftrightarrow\)a2 - 2a + 1 + b2 + 4b + 4 + 4c2 - 4c2 + 1 = 0
\(\Leftrightarrow\)( a - 1 )2 + ( b + 2 )2 + ( 2c - 1 )2 = 0
\(\Leftrightarrow\)\(\hept{\begin{cases}\left(a-1\right)^2=0\\\left(b+2\right)^2=0\\\left(2c-1\right)^2=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}a-1=0\\b+2=0\\2c-1=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}a=1\\b=-2\\c=\frac{1}{2}\end{cases}}\)
Vậy a = 1 , b = -2 , c = \(\frac{1}{2}\)
NGọc Nguyễn suy ra 1 phát luôn là sai đấy mà ko có nhân xét gì à
