- Với \(x=0\) hoặc \(y=0\) đều ko phải là nghiệm của hệ
- Với \(xy\ne0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(2xy+3\right)\left(4x^2y^2-6xy+9\right)=91y^3\\x\left(2xy+3\right)=14y^2\end{matrix}\right.\)
Chia vế cho vế:
\(\dfrac{4x^2y^2-6xy+9}{x}=\dfrac{13y}{2}\)
\(\Rightarrow2\left(4x^2y^2-6xy+9\right)=13xy\)
\(\Leftrightarrow8\left(xy\right)^2-25xy+18=0\)
\(\Rightarrow\left[{}\begin{matrix}xy=2\\xy=\dfrac{9}{8}\end{matrix}\right.\)
Thay vào pt đầu:
TH1: \(\left\{{}\begin{matrix}xy=2\\8\left(xy\right)^3+27=91y^3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{y}\\91y^3=91\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}xy=\dfrac{9}{8}\\8\left(xy\right)^3+27=91y^3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{9}{8y}\\y^3=\dfrac{27}{64}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\)
