\(\left\{{}\begin{matrix}x^2+3xy+y^2=0\left(1\right)\\x^3-y^2=y^3-x^2\left(2\right)\end{matrix}\right.\)
Xét pt(2):\(x^3-y^2=y^3-x^2\)
\(\Leftrightarrow x^3-y^3=y^2-x^2\)
\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)=\left(y-x\right)\left(y+x\right)\)
\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2+x+y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y-2xy\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-y=0\\x+y=2xy\end{matrix}\right.\)(ĐK:\(x\le-y\))
\(\Leftrightarrow\left[{}\begin{matrix}x=y=0\\\left(x+y\right)^2=4x^2y^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=y=0\\-xy=4x^2y^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=y=0\\4xy=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=y=0\\2xy=-\frac{1}{2}=x+y\end{matrix}\right.\)
Với \(x+y=-\frac{1}{2}\Rightarrow x=-\frac{1}{2}-y\)
(1)\(\Leftrightarrow\frac{1}{4}+y^2+y+y^2-\frac{3}{2}y-3y^2=0\)
\(\Leftrightarrow-y^2-\frac{1}{2}y+\frac{1}{4}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=\frac{-1+\sqrt{5}}{4}\\y=\frac{-1-\sqrt{5}}{4}\end{matrix}\right.\)
=>x=...(nhớ đối chiếu với đk)
