Ta có :
\(a^2+b^2-4a+2b+5=0\)
\(\Leftrightarrow\left(a^2-4a+4\right)+\left(b^2+2b+1\right)=0\)
\(\Leftrightarrow\left(a-2\right)^2+\left(b+1\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}a-2=0\\b+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=2\\b=-1\end{matrix}\right.\)
Vậy \(a=2\) và \(b=-1\)
ta có : \(a^2+b^2-4a+2b+5=0\Leftrightarrow\left(a ^2-4a+4\right)+\left(b^2+2b+1\right)=0\)
\(\Leftrightarrow\left(a-2\right)^2+\left(b+1\right)^2=0\)
ta có : \(\left(a-2\right)^2\ge0\forall a\) và \(\left(b+1\right)^2\ge0\forall b\)
\(\Rightarrow\left(a-2\right)^2+\left(b+1\right)^2\ge0\forall a;b\)
\(\Rightarrow\left(a-2\right)^2+\left(b+1\right)^2=0\Leftrightarrow\left\{{}\begin{matrix}\left(a-2\right)^2=0\\\left(b+1\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a-2=0\\b+1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=-1\end{matrix}\right.\) vậy \(a=2;b=-1\)
a2+b2-4a+2b+5=0
⇔a2+b2-4a+2b+1+4=0
⇔(a2-4a+4)+(b2+2b+1)=0
⇔(a-2)2+(b+1)2=0
⇔\(\left\{{}\begin{matrix}a-2=0\\b+1=0\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}a=2\\b=-1\end{matrix}\right.\)
