\(\hept{\begin{cases}x^3y^3+1=2y^3\\\frac{x^2}{y}+\frac{x}{y^2}=2\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x^3+\frac{1}{y^3}=2\\\frac{x}{y}\left(x+\frac{1}{y}\right)=2\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}\left(x+\frac{1}{y}\right)\left(x^2+\frac{x}{y}+\frac{1}{y^2}\right)=2\\\frac{x}{y}\left(x+\frac{1}{y}\right)=2\end{cases}}\)
Suy ra:
\(\left(x+\frac{1}{y}\right)\left(x^2+\frac{x}{y}+\frac{1}{y^2}\right)=\frac{x}{y}\left(x+\frac{1}{y}\right)\)
\(\Leftrightarrow\left(x+\frac{1}{y}\right)\left(x^2+\frac{x}{y}+\frac{1}{y^2}-\frac{x}{y}\right)=0\)
\(\Leftrightarrow\left(x+\frac{1}{y}\right)\left(x^2+\frac{1}{y^2}\right)=0\)
Nhận thấy \(x^2+\frac{1}{y^2}\ne0\) vì nếu \(x^2+\frac{1}{y^2}=0\) thì \(\hept{\begin{cases}x=0\\y=0\end{cases}}\) (vô lý).
Suy ra: \(x+\frac{1}{y}=0\).
vậy đề bài sai.
