Question

Giải hệ phương trình

a) \(\left\{{}\begin{matrix}x+y+xy=-1\\x^2+y^2-3xy=11\end{matrix}\right.\)

b) \(\left\{{}\begin{matrix}2x^2+y=5x\\2y^2+x=5y\end{matrix}\right.\)

Answer 1

\(a,HPT\Leftrightarrow\left\{{}\begin{matrix}x+y+xy=-1\\\left(x+y\right)^2-5xy=11\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+y=S\\xy=P\end{matrix}\right.\), HPTTT:

\(\Leftrightarrow\left\{{}\begin{matrix}S+P=-1\\S^2-5P=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}P=-1-S\\S^2+S-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}S=2\\P=-3\end{matrix}\right.\\\left\{{}\begin{matrix}S=-3\\P=2\end{matrix}\right.\end{matrix}\right.\)

Với \(\left\{{}\begin{matrix}S=2\\P=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=2\\xy=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2-y\\y^2-2y-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=3\end{matrix}\right.\end{matrix}\right.\)

Với \(\left\{{}\begin{matrix}S=-3\\P=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=-3\\xy=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3-y\\y^2+3y+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1\\y=-2\end{matrix}\right.\end{matrix}\right.\)

Vậy HPT có n₀ \(\left(x;y\right)=\left\{\left(3;-1\right);\left(-1;3\right);\left(-2;-1\right);\left(-1;-2\right)\right\}\)