\(x^2+4y^2+13-6x+8y=0\)
\(=\left(x-3\right)^2+4\left(y-1\right)^2-26\ge-26\)
\(Min\) là \(-26\Leftrightarrow x=3;y=1\)
Vậy................
\(x^2+4y^2+13-6x+8y=0\)
\(\Rightarrow x^2+4y^2+9+4-6x+8y=0\)
\(\Rightarrow x^2-6x+9+4y^2+8y+4=0\)
\(\Rightarrow\left(x^2-6x+9\right)+\left(4y^2+8y+4\right)=0\)
\(\Rightarrow\left(x^2-2.x.3+3^2\right)+\left[\left(2y\right)^2+2.2y.2+2^2\right]=0\)
\(\Rightarrow\left(x-3\right)^2+\left(2y+2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-3\right)^2=0\\\left(2y+2\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-3=0\\2y+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0+3\\2y=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)
Vậy \(x=3;y=-1.\)
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