Ta có: \(\left\{{}\begin{matrix}x\left(x-3y\right)=4\left(y^2+2\right)\left(1\right)\\\left(xy-4\right)\left(x+y\right)=8\left(2\right)\end{matrix}\right.\)

\(\left(2\right)\Rightarrow xy-4;x+y\ne0\)

\(\left(1\right)\Leftrightarrow x^2-3xy-4y^2=8\) (*)

Từ (*) và (2) \(\Rightarrow x^2-3xy-4y^2=\left(xy-4\right)\left(x+y\right)\)

\(\Leftrightarrow x\left(x-4y\right)+y\left(x-4y\right)=\left(xy-4\right)\left(x+y\right)\)

\(\Leftrightarrow\left(x+y\right)\left(x-4y\right)=\left(x+y\right)\left(xy-4\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\left(L\right)\\x-4y=xy-4\end{matrix}\right.\) \(\Leftrightarrow x\left(1-y\right)+4\left(1-y\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(1-y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\y=1\end{matrix}\right.\)

x = -4 thay vào (*), ta được: \(16-3.\left(-4\right)y-4y^2=8\)

\(\Leftrightarrow8+12y-4y^2=0\) \(\Leftrightarrow y^2-3y-2=0\) 

\(\Leftrightarrow y=\dfrac{3\pm\sqrt{17}}{2}\) ( dùng \(\Delta\) )

y=1 thay vào (*), ta được: \(x^2-3x-4=8\)

\(\Leftrightarrow x^2-3x-12=0\) \(\Leftrightarrow x=\dfrac{3\pm\sqrt{57}}{2}\)

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