\(\Leftrightarrow\left\{{}\begin{matrix}x-y+2xy=5\\\left(x-y\right)^2+3xy=7\end{matrix}\right.\) \(\Rightarrow\left(5-2xy\right)^2+3xy=7\)
\(\Leftrightarrow4\left(xy\right)^2-22xy+18=0\Rightarrow\left[{}\begin{matrix}xy=1\Rightarrow x-y=3\\xy=\frac{9}{2}\Rightarrow x-y=-4\end{matrix}\right.\)
Đến đây chắc được rồi
Đặt \(\left\{{}\begin{matrix}S=x-y\\P=xy\end{matrix}\right.\)
Ta có: \(\left\{{}\begin{matrix}S+2P=5\\S^2+3P=7\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}1,5S+3P=7,5\\S^2+3P=7\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3P=7,5-1,5S\\S^2+3P=7\end{matrix}\right.\)
\(\Rightarrow S^2-1,5S+7,5=7\Rightarrow S^2-1,5S+0,5=0\)
\(\Rightarrow2S^2-3S+1=0\Rightarrow\left[{}\begin{matrix}S=1\\S=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}S=1\\P=2\end{matrix}\right.\\\left\{{}\begin{matrix}S=\frac{1}{2}\\P=\frac{9}{4}\end{matrix}\right.\end{matrix}\right.\)
Trường hợp \(\left\{{}\begin{matrix}S=1\\P=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=1\\xy=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=-2\end{matrix}\right.\end{matrix}\right.\)
Trường hợp
\(\left\{{}\begin{matrix}S=\frac{1}{2}\\P=\frac{9}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=\frac{1}{2}\\xy=\frac{9}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\frac{\sqrt{37}+1}{4}\\y=\frac{\sqrt{37}-1}{4}\end{matrix}\right.\\\left\{{}\begin{matrix}x=\frac{1-\sqrt{37}}{4}\\y=\frac{-1-\sqrt{37}}{4}\end{matrix}\right.\end{matrix}\right.\)
