Ấy, nhìn không kỹ nên sai sót kỹ thuật rồi, bước đặt nhân tử chung bị nhầm.
Làm lại cho chính xác hơn:
Hệ đã cho tương đương \(\left\{{}\begin{matrix}\left(x^2-2\right)^2-4+\left(y-3\right)^2=0\left(1\right)\\y=\dfrac{22-x^2}{x^2+2}\end{matrix}\right.\)
Đặt \(x^2-2=t\Rightarrow x^2=t+2\Rightarrow y=\dfrac{20-t}{t+4}\Rightarrow y-3=\dfrac{4\left(2-t\right)}{t+4}\left(2\right)\)
Thay (2) vào (1):
\(t^2-4+\dfrac{16\left(2-t\right)^2}{\left(t+4\right)^2}=0\Leftrightarrow\left(t-2\right)\left(t+2\right)+\dfrac{16\left(t-2\right)^2}{\left(t+4\right)^2}=0\)
\(\Leftrightarrow\left(t-2\right)\left(t+2+\dfrac{16\left(t-2\right)}{\left(t+4\right)^2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t-2=0\\t+2+\dfrac{16\left(t-2\right)}{\left(t+4\right)^2}=0\end{matrix}\right.\)
TH1: \(t-2=0\Rightarrow t=2\Rightarrow x^2=4\) \(\Rightarrow\left[{}\begin{matrix}x=-2;y=3\\x=2;y=3\end{matrix}\right.\)
TH2: \(t+2+\dfrac{16\left(t-2\right)}{\left(t+4\right)^2}=0\Leftrightarrow\left(t+2\right)\left(t^2+8t+16\right)+16t-32=0\)
\(\Leftrightarrow t^3+8t^2+16t+2t^2+16t+32+16t-32=0\)
\(\Leftrightarrow t^3+10t^2+48t=0\)
\(\Leftrightarrow t\left(t^2+10t+48\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}t=0\\t^2+10t+48=0\left(vn\right)\end{matrix}\right.\) \(\Rightarrow x^2=2\) \(\Rightarrow\left[{}\begin{matrix}x=-\sqrt{2};y=5\\x=\sqrt{2};y=5\end{matrix}\right.\)
Vậy hệ đã cho có 4 cặp nghiệm:
\(\left(x;y\right)=\left(-2;3\right);\left(2;3\right);\left(-\sqrt{2};5\right);\left(\sqrt{2};5\right)\)
\(\left\{{}\begin{matrix}\left(x^2-2\right)^2-4+\left(y-3\right)^2=0\\\left(x^2+2\right).y=22-x^2\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}\left(x^2-2\right)^2-4+\left(y-3\right)^2=0\\y=\dfrac{22-x^2}{x^2+2}\end{matrix}\right.\)
Đặt \(x^2-2=t\ge-2\)
\(\Rightarrow x^2=t+2\Rightarrow y=\dfrac{20-t}{t+4}\Rightarrow y-3=\dfrac{8-4t}{t+4}=\dfrac{4\left(2-t\right)}{t+4}\)
Thay vào pt trên ta được:
\(t^2-4+\dfrac{16\left(2-t\right)^2}{\left(t+4\right)^2}=0\Leftrightarrow\left(t-2\right)\left(t+2\right)+\dfrac{16\left(t-2\right)^2}{\left(t+4\right)^2}=0\)
\(\Leftrightarrow\left(t-2\right)\left(t+2+\dfrac{16}{\left(t+4\right)^2}\right)=0\)
\(\Leftrightarrow t-2=0\) (do \(t+2+\dfrac{16}{\left(t+4\right)^2}>0\) \(\forall t\ge-2\) )
\(\Rightarrow t=2\Rightarrow x^2-2=2\Rightarrow x^2=4\)
\(\Rightarrow\left[{}\begin{matrix}x=2\Rightarrow y=3\\x=-2\Rightarrow y=3\end{matrix}\right.\)
Vậy hệ đã cho có 2 cặp nghiệm:
\(\left(x;y\right)=\left(-2;3\right);\left(2;3\right)\)