[tex]\left\{\begin{matrix} x^2(y^2+1)+2y(x^2+x+1)=3\\ (x^2+x)(y^2+y)=1 \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x^2(y^2+2y+1)+2y(x+1)=3\\ x(x+1)y(y+1)=1 \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x^2(y+1)^2+2y(x+1)=3\\ x(y+1)y(x+1)=1 \end{matrix}\right.[/tex]
Đặt [tex]a=x(y+1),b=y(x+1)[/tex]
Hệ đã cho trở thành:[tex]\left\{\begin{matrix} a^2+2b=3\\ ab=1 \end{matrix}\right.\Rightarrow \left\{\begin{matrix} a^2+2b-3=0\\ b=\frac{1}{a} \end{matrix}\right.\Rightarrow \left\{\begin{matrix} a^2+\frac{2}{a}-3=0\\ b=\frac{1}{a} \end{matrix}\right.\Rightarrow \left\{\begin{matrix} a^3-3a+2=0\\ b=\frac{1}{a} \end{matrix}\right.\Rightarrow \left\{\begin{matrix} (a-1)(a^2+a-2)=0\\ b=\frac{1}{a} \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} (a-1)^2(a+2)=0\\ b=\frac{1}{a} \end{matrix}\right.[/tex]
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