\(\hept{\begin{cases}x^4y^2+y+1=x^2y\left(y+2\right)\left(1\right)\\x^3y-\frac{x}{y}+x=2xy\left(2\right)\end{cases}}\)
\(\left(1\right)\Leftrightarrow\left(x^4y^2-x^2y\right)+\left(-x^2y^2+y\right)+\left(-x^2y+1\right)=0\)
\(\Leftrightarrow\left(x^2y-1\right)\left(x^2y-y-1\right)=0\)
Thế vô PT (2) giải tiếp