\(\Leftrightarrow\left\{{}\begin{matrix}x^2+1=-xy-y^2+4y\\\left(x^2+1\right)\left(x+y-2\right)=y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+1=-y\left(x+y-4\right)\\\left(x^2+1\right)\left(x+y-2\right)=y\end{matrix}\right.\)
Thế trên xuống dưới:
\(-y\left(x+y-4\right)\left(x+y-2\right)=y\)
\(\Leftrightarrow\left[{}\begin{matrix}y=0\Rightarrow x^2+1=0\left(l\right)\\\left(x+y-4\right)\left(x+y-2\right)=-1\left(1\right)\end{matrix}\right.\)
Xét (1), đặt \(x+y-4=t\)
\(\Rightarrow t\left(t+2\right)=-1\Leftrightarrow\left(t+1\right)^2=0\Rightarrow t=-1\)
\(\Rightarrow x+y-4=-1\Rightarrow y=3-x\)
Thế vào pt đầu:
\(x^2+x\left(3-x\right)+1=4\left(3-x\right)-\left(3-x\right)^2\)
\(\Leftrightarrow...\)