\(\Leftrightarrow\left\{{}\begin{matrix}1+\frac{10xy}{\left(x^2+3\right)\left(y^2+1\right)}=0\\\frac{x}{x^2+3}+\frac{y}{y^2+1}=-\frac{3}{20}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\frac{x}{\left(x^2+3\right)}.\frac{y}{\left(y^2+1\right)}=-\frac{1}{10}\\\frac{x}{x^2+3}+\frac{y}{y^2+1}=-\frac{3}{20}\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\frac{x}{x^2+3}=a\\\frac{y}{y^2+1}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}ab=-\frac{1}{10}\\a+b=-\frac{3}{20}\end{matrix}\right.\)
Theo Viet đảo, a và b là nghiệm:
\(t^2+\frac{3}{20}t-\frac{1}{10}=0\Rightarrow\left[{}\begin{matrix}t=\frac{1}{4}\\t=-\frac{2}{5}\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}\frac{x}{x^2+3}=\frac{1}{4}\\\frac{y}{y^2+1}=-\frac{2}{5}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x^2-4x+3=0\\2y^2+5y+2=0\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}\frac{x}{x^2+3}=-\frac{2}{5}\\\frac{y}{y^2+1}=\frac{1}{4}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x^2+5x+6=0\\y^2-4y+1=0\end{matrix}\right.\)
