\(a^2-6a+6b+b^2=-10\)
\(\Leftrightarrow a^2-2a+6b+b^2+10=0\)
\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2+6b+9\right)=0\)
\(\Leftrightarrow\left(a^2-2.a.1+1^2\right)+\left(b^2+2.b.3+3^2\right)=0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b+3\right)^2=0\) (1)
Vì \(\left(a-1\right)^2+\left(b+3\right)^2\ge0\) với mọi a;b
Nên để thỏa mãn (1) thì \(\left(a-1\right)^2=\left(b+3\right)^2=0\Leftrightarrow a=1;b=-3\)