ĐKXĐ: \(y\ne0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+\dfrac{1}{y}\right)^2-\dfrac{x}{y}=3\\x+\dfrac{1}{y}+\dfrac{x}{3}=3\end{matrix}\right.\)
Cộng vế với vế:
\(\left(x+\dfrac{1}{y}\right)^2+\left(x+\dfrac{1}{y}\right)=6\)
\(\Leftrightarrow\left(x+\dfrac{1}{y}\right)^2+\left(x+\dfrac{1}{y}\right)-6=0\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{y}=2\\x+\dfrac{1}{y}=-3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=2-x\\\dfrac{1}{y}=-3-x\end{matrix}\right.\)
Thế xuống pt dưới:
\(\Rightarrow\left[{}\begin{matrix}2+x\left(2-x\right)=3\\-3+x\left(-3-x\right)=3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-2x+1=0\\x^2+3x+6=0\left(vô-nghiệm\right)\end{matrix}\right.\)
\(\Rightarrow x=1\Rightarrow y=1\)
