\(\left\{{}\begin{matrix}x^2+\left(y+\dfrac{1}{y}\right)^2=5\\x\left(y+\dfrac{1}{y}\right)=2\end{matrix}\right.\)
Đặt \(y+\dfrac{1}{y}=a\) \(\Rightarrow\left\{{}\begin{matrix}x^2+a^2=5\\x.a=2\Rightarrow a=\dfrac{2}{x}\end{matrix}\right.\)
\(\Rightarrow x^2+\left(\dfrac{2}{x}\right)^2=5\Leftrightarrow x^4-5x^2+4=0\) \(\Rightarrow\left[{}\begin{matrix}x^2=4\\x^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pm2\\x=\pm1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}a=\pm1\\a=\pm2\end{matrix}\right.\)
\(a=1\Rightarrow y+\dfrac{1}{y}=1\Rightarrow y^2-y+1=0\) (vô nghiệm)
\(a=-1\Rightarrow y+\dfrac{1}{y}=-1\Rightarrow y^2+y+1=0\) (vô nghiệm)
\(a=2\Rightarrow y+\dfrac{1}{y}=2\Rightarrow y^2-2y+1=0\Rightarrow y=1\)
\(a=-2\Rightarrow y+\dfrac{1}{y}=-2\Rightarrow y^2+2y+1=0\Rightarrow y=-1\)
Vậy hệ đã cho có 2 cặp nghiệm:
\(\left(x;y\right)=\left(1;1\right);\left(-1;-1\right)\)
