\(\left\{{}\begin{matrix}x^2\left(y^2+1\right)+2y\left(x^2+x+1\right)=3\\\left(x^2+x\right)\left(y^2+y\right)=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2\left(y+1\right)^2+2y\left(x+1\right)=3\\xy\left(x+1\right)\left(y+1\right)=1\end{matrix}\right.\)
Đặt \(x\left(y+1\right)=a\text{ và }y\left(x+1\right)=b\)
\(\Rightarrow\left\{{}\begin{matrix}a^2+2b=3\\ab=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2+\dfrac{2}{a}-3=0\left(1\right)\\b=\dfrac{1}{a}\end{matrix}\right.\) (a ≠ 0)
\(\left(1\right)\Leftrightarrow a^3-3a+2=0\)
\(\Leftrightarrow\left(a-1\right)^2\left(a+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a=-2\end{matrix}\right.\left(\text{nhận}\right)\Rightarrow\left[{}\begin{matrix}b=1\\b=-\dfrac{1}{2}\end{matrix}\right.\)
➤ TH1: \(x\left(y+1\right)=y\left(x+1\right)=1\)
\(\Rightarrow\left\{{}\begin{matrix}xy+x=yx+y\\x\left(y+1\right)=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y\\y^2+y-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=y=\dfrac{-1+\sqrt{5}}{2}\\x=y=\dfrac{-1-\sqrt{5}}{2}\end{matrix}\right.\)
➤ TH2: \(\left\{{}\begin{matrix}x\left(y+1\right)=-2\left(1\right)\\y\left(x+1\right)=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow x-y=-\dfrac{3}{2}\)
\(\text{Thay }x=y-\dfrac{3}{2}\text{ vào }\left(1\right)\Rightarrow\left(y-\dfrac{3}{2}\right)\left(y+1\right)=-2\)
\(\Leftrightarrow2y^2-y+1=0\)
Ta có: \(\Delta=\left(-1\right)^2-4.2.1=-7< 0\)
⇒ Phương trình vô nghiệm.
Vậy hệ phương trình có nghiệm \(\left(x;y\right)=\left(\dfrac{-1+\sqrt{5}}{2};\dfrac{-1+\sqrt{5}}{2}\right);\left(\dfrac{-1-\sqrt{5}}{2};\dfrac{-1-\sqrt{5}}{2}\right)\)
