\(\left\{{}\begin{matrix}\left(2x^2-3x+4\right)\left(2y^2-3y+4\right)=18\\x^2+y^2+xy-7x-6y+14=13\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x^3+y^3=91\\4x^2+3y^2=16x+9y\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x-\frac{1}{x^3}=y-\frac{1}{y^3}\\\left(x-4y\right)\left(2x-y+\text{4}\right)=-36\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x^2+y^2=2xy+1\\x^5+y^3+1=0\end{matrix}\right.\)

Giải hpt;\(\left\{{}\begin{matrix}x^3+y^3=1\\\left(x+y\right)^2y=2\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x^2+y^2=\frac{1}{2}\\2x^3+6xy^2=1\end{matrix}\right.\)