Ta có : \(\left\{{}\begin{matrix}x^2y^2-xy-2=0\\x+y=x^2y^2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\left(xy-\frac{1}{2}\right)^2=\frac{9}{4}\\x+y=x^2y^2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}xy-\frac{1}{2}=\frac{3}{2}\\xy-\frac{1}{2}=-\frac{3}{2}\end{matrix}\right.\\x+y=x^2y^2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}xy=2\\xy=-1\end{matrix}\right.\\x+y=x^2y^2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\frac{2}{y}\\x=-\frac{1}{y}\end{matrix}\right.\\\left[{}\begin{matrix}\frac{2}{y}+y=4\\y-\frac{1}{y}=1\end{matrix}\right.\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\frac{2}{y}\\x=-\frac{1}{y}\end{matrix}\right.\\\left[{}\begin{matrix}2+y^2=4y\\y^2-1=y\end{matrix}\right.\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\frac{2}{y}\\x=-\frac{1}{y}\end{matrix}\right.\\\left[{}\begin{matrix}\left(y-2\right)^2=2\\\left(y-\frac{1}{2}\right)^2=\frac{5}{4}\end{matrix}\right.\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\frac{2}{y}\\x=-\frac{1}{y}\end{matrix}\right.\\\left[{}\begin{matrix}y=\sqrt{2}+2\\y=2-\sqrt{2}\\y=\frac{1+\sqrt{5}}{2}\\y=\frac{1-\sqrt{5}}{2}\end{matrix}\right.\end{matrix}\right.\)
- THay y vào phương trình x ta được :
\(\left[{}\begin{matrix}x=\frac{2}{\sqrt{2}+2}=2-\sqrt{2}\\x=\frac{2}{2-\sqrt{2}}=2+\sqrt{2}\\y=\frac{1-\sqrt{5}}{2}\\y=\frac{1+\sqrt{5}}{2}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^2y^2-xy-2=0\\x+y=x^2y^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(xy\right)^2-xy-2=0\\x+y=\left(xy\right)^2\end{matrix}\right.\) (I)
Đặt \(\left\{{}\begin{matrix}xy=a\\x+y=b\end{matrix}\right.\)
(I) \(\Leftrightarrow\left\{{}\begin{matrix}a^2-a-2=0\\a=b^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}a=2\\a=-1\end{matrix}\right.\\b^2=a\end{matrix}\right.\)\(\Leftrightarrow\) \(\left[{}\begin{matrix}\left\{{}\begin{matrix}a=2\\b=\sqrt{2}\end{matrix}\right.\\\left\{{}\begin{matrix}a=-1\\b^2=a\end{matrix}\right.\end{matrix}\right.\)
Ta có \(\left\{{}\begin{matrix}a=-1\\b^2=a\end{matrix}\right.\)( Vô nghiệm)
Lại có: \(\left\{{}\begin{matrix}a=2\\b=\sqrt{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}xy=2\\x+y=\sqrt{2}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{2}-y\\\left(\sqrt{2}-y\right)y=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{2}-y\\\sqrt{2}y-y^2-2=0\end{matrix}\right.\) (Vô nghiệm)
Vậy hpt trên vô nghiệm
sai đề bài
sửa lại chỗ x+y=x2y2 thành x2+y2=x2y2
