\(5x^2+2y^2-6xy+16x-8y+16=0\)
\(\Rightarrow10x^2+4y^2-12xy+32x-16y+32=0\)
\(\Rightarrow\left(9x^2-12xy+4y^2\right)+\left(24x-16y\right)+16+\left(x^2+8x+16\right)=0\)
\(\Rightarrow\left(3x-2y\right)^2+2.\left(3x-2y\right).4+4^2+\left(x+4\right)^2=0\)
\(\Rightarrow\left(3x-2y+4\right)^2+\left(x+4\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}3x-2y+4=0\\x+4=0\end{cases}\Rightarrow}\hept{\begin{cases}-12-2y+4=0\\x=-4\end{cases}\Rightarrow\hept{\begin{cases}y=-4\\x=-4\end{cases}}}\)
Vậy \(x=y=-4\)
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