Question

giải hệ pt sau: \(\left\{{}\begin{matrix}xy^2+y=-6x^2\\x^3y^3+1=19x^3\end{matrix}\right.\)

Answer 1

\(\Leftrightarrow\left\{{}\begin{matrix}y\left(xy+1\right)=-6x^2\\\left(xy+1\right)\left(x^2y^2-xy+1\right)=19x^3\end{matrix}\right.\)

Nhận thấy \(x=0\) ko phải nghiệm, chia vế cho vế:

\(\frac{y}{x^2y^2-xy+1}=\frac{-6}{19x}\)

\(\Leftrightarrow-19xy=6x^2y^2-6xy+6\)

\(\Leftrightarrow6x^2y^2+13xy+6=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy=-\frac{2}{2}\\xy=-\frac{3}{2}\end{matrix}\right.\)

Thay xuống pt dưới ...