Question

\(\left\{{}\begin{matrix}x^2y+2x+3y=6\\3xy+x+y=5\end{matrix}\right.\)

Answer 1

Trừ vế cho vế:

\(x^2y-3xy+x+2y-1=0\)

\(\Leftrightarrow xy\left(x-1\right)-2y\left(x-1\right)+x-1=0\)

\(\Leftrightarrow\left(x-1\right)\left(xy-2y+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\Rightarrow y=...\\xy-2y+1=0\left(1\right)\end{matrix}\right.\)

Kết hợp (1) với pt ban đầu: \(\left\{{}\begin{matrix}3xy+x+y=5\\xy-2y+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3xy+x+y=5\\3xy-6y+3=0\end{matrix}\right.\) \(\Rightarrow x+7y-8=0\Rightarrow x=8-7y\)

Thế vào (1): \(y\left(8-7y\right)-2y+1=0\Leftrightarrow-7y^2+6y+1=0\Leftrightarrow...\)