ĐK \(x,y\ne0\)
lấy dưới trừ trên ta đc \(3x-3y=\dfrac{x^2+2}{y^2}-\dfrac{y^2+2}{x^2}\)
\(\Leftrightarrow3x-3y=\dfrac{x^4+2x^2-y^4-2y^2}{x^2y^2}\)
\(\Leftrightarrow3x^2y^2\left(x-y\right)=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)+2\left(x-y\right)\left(x+y\right)\)
\(\Leftrightarrow3x^2y^2\left(x-y\right)-\left(x-y\right)\left(x+y\right)\left(x^2+y^2+2\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left[3x^2y^2-\left(x+y\right)\left(x^2+y^2+2\right)\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=y\left(1\right)\\3x^2y^2=\left(x+y\right)\left(x^2+y^2+2\right)\left(2\right)\end{matrix}\right.\)
TH1 \(\left\{{}\begin{matrix}x=y\\3x=\dfrac{x^2+2}{y^2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y\\3x^3-x^2-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
TH2 \(\left\{{}\begin{matrix}\left(x+y\right)\left(x^2+y^2+2\right)=3x^2y^2\\3x=\dfrac{x^2+2}{y^2}\end{matrix}\right.\)
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