Nhận thấy \(y=3\) không phải là nghiệm
\(x\left(y-3\right)=9y+1\Rightarrow x=\frac{9y+1}{y-3}\)
Thay xuống dưới:
\(\left(\frac{9y+1}{y-3}-1\right)^2y^2+2y+1=0\)
\(\Leftrightarrow\frac{16\left(2y+1\right)^2y^2}{\left(y-3\right)^2}+2y+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2y+1=0\Rightarrow...\\\frac{16y^2\left(2y+1\right)}{\left(y-3\right)^2}+1=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow32y^3+17y^2-6y+9=0\)
\(\Leftrightarrow\left(y+1\right)\left(32y^2-15y+9\right)=0\)