Question

giải hệ

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

Answer 1

ĐKXĐ: \(xy\ne0\)

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

Chia vế cho vế:

\(\Rightarrow\frac{x}{y}=\frac{y^2+2}{x^2+2}\)

\(\Leftrightarrow x^3+2x=y^3+2y\)

\(\Leftrightarrow x^3-y^3+2\left(x-y\right)=0\)

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

\(\Leftrightarrow x=y\)

Thay vào pt đầu: \(3x=\frac{x^2+2}{x^2}\Leftrightarrow3x^3-x^2-2=0\)

\(\Leftrightarrow\left(x-1\right)\left(3x^2+2x+2\right)=0\)