Question

cho hệ phương trình:

\(\left\{{}\begin{matrix}x^2+y^2=5\\x^4-x^2y^2+y^4=13\end{matrix}\right.\)

Answer 1

\(\left\{{}\begin{matrix}x^2+y^2=5\\\left(x^2+y^2\right)^2-3x^2y^2=13\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=5\\x^2y^2=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y^2=5-x^2\\x^2y^2-4=0\end{matrix}\right.\)

\(\Rightarrow x^2\left(5-x^2\right)-4=0\Leftrightarrow-x^4+5x^2-4=0\)

\(\Rightarrow\left[{}\begin{matrix}x^2=1\Rightarrow y^2=4\\x^2=4\Rightarrow y^2=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\pm1;y=\pm2\\x=\pm2;y=\pm1\end{matrix}\right.\)