Question

giải hệ phương trình :

x^2 + y^2 = 5

x^4 + y^4 - x^2y^2 = 13

Answer 1

Đặt \(\left\{{}\begin{matrix}S=x^2+y^2\\P=x^2y^2\end{matrix}\right.\) khi đó ta có hệ:

\(\left\{{}\begin{matrix}S=5\\S^2-3P=13\end{matrix}\right.\)\(\Rightarrow25-3P=13\Rightarrow3P=12\Rightarrow P=4\)

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

\(\Rightarrow-\left(y-2\right)\left(y+2\right)\left(y-1\right)\left(y+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}y=\pm2\\y=\pm1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\mp1\\x=\mp2\end{matrix}\right.\)