Question

Giải hệ phương trình: \(\left\{{}\begin{matrix}x^3y+2xy^3-x^2-2y^2+xy=1\\\frac{x}{x^4+y^2}+\frac{y}{y^4+x^2}=1\end{matrix}\right.\)

Answer 1

\(x^2\left(xy-1\right)+2y^2\left(xy-1\right)+\left(xy-1\right)=0\)

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

\(\Leftrightarrow xy=1\) (do \(x^2+2y^2+1>0\)) \(\Rightarrow y=\frac{1}{x}\)

Thay vào pt dưới:

\(\frac{x}{x^4+\frac{1}{x^2}}+\frac{\frac{1}{x}}{\frac{1}{x^4}+x^2}=1\Leftrightarrow\frac{x^3}{x^6+1}+\frac{x^3}{x^6+1}=1\Leftrightarrow\frac{2x^3}{x^6+1}=1\)

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