Ta có : \(5x^2+2y^2-6xy+16x-8y+16=0\)
\(\Leftrightarrow10x^2+4y^2-12xy+32x-16y+32=0\)
\(\Leftrightarrow\left(4y^2-2.2y.3x+9x^2\right)+x^2+32x-16y+32=0\)
\(\Leftrightarrow\left(2y-3x\right)^2-2.4.\left(2y-3x\right)+16+x^2+8x+16=0\)
\(\Leftrightarrow\left(2y-3x-4\right)^2+\left(x+4\right)^2=0\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(2y-3x-4\right)^2=0\\\left(x+4\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=-4\end{matrix}\right.\)
Vậy \(\left(x,y\right)=\left(-4,4\right)\)
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