\(a^2-2a+6b+b^2=-10\)
\(\Rightarrow a^2-2a+6b+b^2+10=0\)
\(\Rightarrow\left(a^2-2a+1\right)+\left(b^2+6b+9\right)=0\)
\(\Rightarrow\left(a-1\right)^2+\left(b+3\right)^2=0\)
Vì \(\left\{{}\begin{matrix}\left(a-1\right)^2\ge0\\\left(b+3\right)^2\ge0\end{matrix}\right.\) với mọi a,b
=> \(\left(a-1\right)^2+\left(b+3\right)^2\ge0\) với mọi a,b
Mà \(\left(a-1\right)^2+\left(b+3\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(a-1\right)^2=0\\\left(b+3\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a-1=0\\b+3=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=1\\b=-3\end{matrix}\right.\)
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