\(\Leftrightarrow\left\{{}\begin{matrix}\left(\frac{x}{y}-2\right)\left(\left(\frac{x}{y}\right)^2+3\left(\frac{x}{y}\right)+6\right)=0\\\left(xy-2\right)\left(xy+3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x}{y}=2\\\left[{}\begin{matrix}xy=2\\xy=-3\end{matrix}\right.\end{matrix}\right.\) (loại \(xy=-3\))
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=2y\\xy=2\end{matrix}\right.\) \(\Rightarrow2y^2=2\Rightarrow\left[{}\begin{matrix}y=1\Rightarrow x=2\\y=-1\Rightarrow x=-2\end{matrix}\right.\)
+ ĐKXĐ : \(y\ne0\)
+ Dễ thấy x = 0 ko là nghiệm của hpt đã cho
+ \(\left(xy\right)^2+xy=6\Rightarrow\left(xy\right)^2+xy-6=0\)
\(\Rightarrow\left(xy-2\right)\left(xy+3\right)=0\)\(\Rightarrow\left[{}\begin{matrix}xy=2\\xy=-3\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}y=\frac{2}{x}\\y=-\frac{3}{x}\end{matrix}\right.\)
+ \(\left(\frac{x}{y}\right)^3+\left(\frac{x}{y}\right)^2=12\Leftrightarrow\left(\frac{x}{y}\right)^3+\left(\frac{x}{y}\right)^2-12=0\)
\(\Leftrightarrow\left(\frac{x}{y}-2\right)\left[\left(\frac{x}{y}\right)^2+\frac{3x}{y}+6\right]=12\)
\(\Leftrightarrow\frac{x}{y}-2=0\) ( do \(\left(\frac{x}{y}\right)^2+\frac{3x}{y}+6>0\forall y\ne0\) )
\(\Leftrightarrow\frac{x}{y}=2\)
Xét 2 TH là được
