Question

Giải hệ:

\(\left\{{}\begin{matrix}x^2+y^2+xy=7\\x^4+y^4+x^2y^2=21\end{matrix}\right.\)

Answer 1

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-xy=7\\\left[\left(x+y\right)^2-2xy\right]^2-x^2y^2=21\end{matrix}\right.\)

Đặt S=x+y; P=xy(\(S^2\ge4P\))

\(\Rightarrow\left\{{}\begin{matrix}S^2-P=7\\\left[S^2-2P\right]^2-P^2=21\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}S^2-P=7\left(1\right)\\\left[7-P\right]^2-P^2=21\left(2\right)\end{matrix}\right.\)

Ta có: \(\left(2\right)\Leftrightarrow P^2-14P+49-P^2=21\)\(\Leftrightarrow P=2\)(TM).

Thay vào (1) ta được: \(S=\pm3\)(TM).

-TH1: ​\(\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\end{matrix}\right.\)

-TH2: \(\left\{{}\begin{matrix}x+y=-3\\xy=2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-1\\y=-2\end{matrix}\right.\\\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\end{matrix}\right.\)

Vậy hpt có nghiệm là (2;1);(1;2);(-1;-2);(-2;-1).