Question

\(\left\{{}\begin{matrix}xy+x+1=7y\\x^2y^2+xy+1=13y^2\end{matrix}\right.\). Giải hệ

Answer 1

- Với \(y=0\) không phải nghiệm

- Với \(y\ne0\)

\(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{x}{y}+\dfrac{1}{y}=7\\x^2+\dfrac{x}{y}+\dfrac{1}{y^2}=13\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{y}+\dfrac{x}{y}=7\\\left(x+\dfrac{1}{y}\right)^2-\dfrac{x}{y}=13\end{matrix}\right.\)

\(\Rightarrow\left(x+\dfrac{1}{y}\right)^2+x+\dfrac{1}{y}-20=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{y}=4\\x+\dfrac{1}{y}=-5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=4-\dfrac{1}{y}\\x=-5-\dfrac{1}{y}\end{matrix}\right.\)

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