\(4x^2+y^2-4x-6y+10=0\)
\(\Rightarrow\left(4x^2-4x+1\right)+\left(y^2-6y+9\right)=0\)
\(\Rightarrow\left(2x-1\right)^2+\left(y-3\right)^2=0\)
Ta có: Với mọi \(x;y\in R\) thì \(\left(2x-1\right)^2+\left(y-3\right)^2\ge0\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=3\end{matrix}\right.\)