\(\left\{{}\begin{matrix}\left(x^2-2\right)^2+\left(y-3\right)^2=4\\\left(y+1\right)\left(x^2+2\right)=24\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\left(x^2-2\right)^2+\left(y-3\right)^2=4\\y-3=\frac{24}{x^2+2}-4\end{matrix}\right.\)
\(\Rightarrow\left(x^2-2\right)^2+\left(\frac{24}{x^2+2}-4\right)^2=4\)
Đặt \(a=x^2+2\)
\(\Rightarrow\left(a-4\right)^2+\left(\frac{24}{a}-4\right)^2=4\)
\(\Leftrightarrow a^4-8a^3+28a^2-192a+576=0\)
\(\Leftrightarrow\left(a^2-10a+24\right)\left(a^2+2a+24\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=6;y=3\\a=4;y=5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x^2=4;y=3\\x^2=2;y=5\end{matrix}\right.\)
đến đây có thể kết luận nghiệm rồi ạ
