Question

giải hệ phương trình sau

\(\left\{{}\begin{matrix}x=\frac{2y}{1-y^2}\\y=\frac{2x}{1-x^2}\end{matrix}\right.\)

Answer 1

ĐKXĐ: \(x;y\ne\pm1\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x-x^3=2x\\x-x\left(\frac{3}{x}\right)^2=\frac{6}{x}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x\left(x^2+1\right)=0\\x^2=15\end{matrix}\right.\) \(\Rightarrow...\)