\(\left\{{}\begin{matrix}x^2y^2-xy-2=0\left(1\right)\\x^2+y^2=x^2y^2\left(2\right)\end{matrix}\right.\)\(đặt:xy=t\Rightarrow\left(1\right)\Leftrightarrow t^2-t-2=0\Leftrightarrow\left[{}\begin{matrix}t=2=xy\\t=-1=xy\end{matrix}\right.\)
\(với:xy=2\Rightarrow\left(2\right)\Leftrightarrow x^2+y^2=4\Leftrightarrow\left(x+y\right)^2-2xy=4\Leftrightarrow\left(x+y\right)^2=8\Leftrightarrow x+y=\pm2\sqrt{2}\)
\(\Rightarrow\left\{{}\begin{matrix}xy=2\\x+y=\pm2\sqrt{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=y=\sqrt{2}\\x=y=-\sqrt{2}\end{matrix}\right.\)
\(với:xy=-1\Rightarrow\left(2\right)\) \(vô\) \(nghiệm\) \(do:VT,VP>0\)
\(\Rightarrow\left(x;y\right)=\left\{\pm\sqrt{2}\right\}\)
{x2y2−xy−2=0(1)x2+y2=x2y2(2){x2y2−xy−2=0(1)x2+y2=x2y2(2)đặt:xy=t⇒(1)⇔t2−t−2=0⇔[t=2=xyt=−1=xyđặt:xy=t⇒(1)⇔t2−t−2=0⇔[t=2=xyt=−1=xy
với:xy=2⇒(2)⇔x2+y2=4⇔(x+y)2−2xy=4⇔(x+y)2=8⇔x+y=±2√2với:xy=2⇒(2)⇔x2+y2=4⇔(x+y)2−2xy=4⇔(x+y)2=8⇔x+y=±22
⇒{xy=2x+y=±2√2⇒{xy=2x+y=±22⇔[x=y=√2x=y=−√2
