Ta có: \(9x^2+8y^2-12xy+6x-16y+10=0\)
\(\Rightarrow9x^2+8y^2-12xy+6x-16y=-10\)
\(=9x^2+2\left(4y^2-6xy+3x-8y\right)=-10\)
\(=9x^2+2\left[3x-6xy+4y\left(y-2\right)\right]\)
\(=9x^2+2\left[3x\left(1-2y\right)+4y\left(y-2\right)\right]\)
\(\Rightarrow\left\{{}\begin{matrix}9x^2=0\\\left\{{}\begin{matrix}1-2y=0\\y-2=0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\\left\{{}\begin{matrix}y=\dfrac{1}{2}\\y=2\end{matrix}\right.\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=0\\\left\{{}\begin{matrix}y=\dfrac{1}{2}\\y=2\end{matrix}\right.\end{matrix}\right.\)
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