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

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

Đặt \(\left\{{}\begin{matrix}x^2+y^2=a>0\\xy=b\end{matrix}\right.\) với \(a\ge2b\)

Ta được: \(\left\{{}\begin{matrix}a+b=7\\a^2-b^2=21\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=7\\\left(a-b\right)\left(a+b\right)=21\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=7\\a-b=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=5\\b=2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x^2+y^2=5\\xy=2\end{matrix}\right.\) \(\Rightarrow x^2+\left(\frac{2}{x}\right)^2=5\Leftrightarrow x^4-5x^2+4=0\)