\(\left\{{}\begin{matrix}xy+3y^2+x=3\\x^2+xy-2y^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(y+1\right)+3y^2-3=0\\x^2+xy-2y^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(y+1\right)+3\left(y+1\right)\left(y-1\right)=0\\x^2+xy-2y^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(y+1\right)\left(x+3y-3\right)=0\\x^2+xy-2y^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}y=-1\\x=3-3y\end{matrix}\right.\\x^2+xy-2y^2=0\end{matrix}\right.\)
+) \(\left\{{}\begin{matrix}y=-1\\x^2+x\cdot\left(-1\right)-2\cdot\left(-1\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x^2-x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-1\\\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left(x;y\right)=\left\{\left(2;-1\right);\left(-1;-1\right)\right\}\)
+) \(\left\{{}\begin{matrix}x=3-3y\\\left(3-3y\right)^2+\left(3-3y\right)\cdot y-2y^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3-3y\\9-18y+9y^2+3y-3y^2-2y^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3-3y\\4y^2-15y+9=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3-3y\\\left[{}\begin{matrix}y=3\\y=\dfrac{3}{4}\end{matrix}\right.\end{matrix}\right.\)
Với \(y=3\Rightarrow x=-6\)
Với \(y=\dfrac{3}{4}\Rightarrow x=\dfrac{3}{4}\)
Vậy: \(\left(x;y\right)=\left\{\left(2;-1\right);\left(-1;-1\right);\left(3;-6\right);\left(\dfrac{3}{4};\dfrac{3}{4}\right)\right\}\)
