Question

Giai hệ phương trình: \(\left\{{}\begin{matrix}x^2+xy+1=4y-y^2\\\left(x^2+1\right)\left(x+y-2\right)=y\end{matrix}\right.\)

Answer 1

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+1=-xy-y^2+4y\\\left(x^2+1\right)\left(x+y-2\right)=y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+1=-y\left(x+y-4\right)\\\left(x^2+1\right)\left(x+y-2\right)=y\end{matrix}\right.\)

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:

\(x^2+x\left(3-x\right)+1=4\left(3-x\right)-\left(3-x\right)^2\)

\(\Leftrightarrow...\)