\(\left\{{}\begin{matrix}x+y+2xy=7\\x^2+y^2=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+2xy=7\\\left(x+y\right)^2-2xy=5\end{matrix}\right.\)
Đặt \(x+y=a;xy=b\)
hpt \(\Leftrightarrow\left\{{}\begin{matrix}a+2b=7\\a^2-2b=5\end{matrix}\right.\)
Cộng theo vế 2 pt ta được:
\(a^2+a-12=0\)
\(\Leftrightarrow\left(a+4\right)\left(a-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-4\\a=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y=-4\\x+y=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y=-4\\xy=\frac{11}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x;y\right)\in\varnothing\\\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\\\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x;y\right)=\left(1;2\right)\\\left(x;y\right)=\left(2;1\right)\end{matrix}\right.\)
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