\(\left\{{}\begin{matrix}2x^2+3xy-2y^2-5\left(2x-y\right)=0\\x^2-2xy-3y^2+15=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y\right)\left(x+2y\right)-5\left(2x-y\right)=0\\x^2-2xy-3y^2+15=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y\right)\left(x+2y-5\right)=0\left(1\right)\\x^2-2xy-3y^2+15=0\left(2\right)\end{matrix}\right.\)
\(PT\left(1\right)\Leftrightarrow\left[{}\begin{matrix}2x-y=0\\x+2y-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{y}{2}\\x=5-2y\end{matrix}\right.\)
Với \(x=\dfrac{y}{2}\) : \(PT\left(2\right)\Leftrightarrow\dfrac{y^2}{4}-y^2-3y^2+15=0\)
\(\Leftrightarrow-15y^2+60=0\)
\(\Leftrightarrow y^2-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=-2\\y=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
Với \(x=5-2y\) : \(PT\left(2\right)\Leftrightarrow\left(5-2y\right)^2-2y\left(5-2y\right)-3y^2+15=0\)
\(\Leftrightarrow4y^2-20y+25+4y^2-10y-3y^2+15=0\)
\(\Leftrightarrow5y^2-30y+40=0\)
\(\Leftrightarrow y^2-6y+8=0\)
\(\Leftrightarrow\left(y-2\right)\left(y-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=2\\y=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
Vậy phương trình có 3 cặp nghiệm : \(\left[{}\begin{matrix}\left(x;y\right)=\left(-1;-2\right)\\\left(x;y\right)=\left(1;2\right)\\\left(x;y\right)=\left(-3;4\right)\end{matrix}\right.\)