Question

[yeens]

giải hệ: \(\left\{{}\begin{matrix}y\left(xy-2\right)=3x^2\\y^2+x^2y+2x=0\end{matrix}\right.\)

Answer 1

[Nguyễn Việt Lâm]

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

Nhận thấy \(\left(x;y\right)=\left(0;0\right)\) là 1 nghiệm

Với \(x;y\ne0\)

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

Nhân vế với vế: \(\left(xy-2\right)\left(xy+2\right)=-3xy\)

\(\Leftrightarrow\left(xy\right)^2+3xy-4=0\Rightarrow\left[{}\begin{matrix}xy=1\\xy=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}y=\dfrac{1}{x}\\y=-\dfrac{4}{x}\end{matrix}\right.\)

Thế vào pt dưới:

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{x^2}+x+2x=0\\\dfrac{16}{x^2}-4x+2x=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}3x^3=-1\\x^3=8\end{matrix}\right.\) \(\Leftrightarrow...\)