\(\Leftrightarrow x^2y+3x^2=y\left(x^4-6x^2+9\right)\)
\(\Leftrightarrow y\left(x^4-7x^2+9\right)=3x^2\)
\(\Leftrightarrow y=\frac{3x^2}{x^4-7x^2+9}\) (do \(x^4-7x^2+9=0\) ko có nghiệm nguyên nên chia thoải mái)
Do y nguyên dương \(\Rightarrow y\ge1\Rightarrow\left\{{}\begin{matrix}x^4-7x^2+9>0\\\frac{3x^2}{x^4-7x^2+9}\ge1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x^4-7x^2+9\ge0\\3x^2\ge x^4-7x^2+9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^4-7x^2+9\ge0\\x^4-10x^2+9\le0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^4-7x^2+9\ge0\\\left(x^2-1\right)\left(x^2-9\right)\le0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^4-7x^2+9\ge0\left(1\right)\\1\le x^2\le9\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Rightarrow x^2=\left\{1;4;9\right\}\)
Thay vào (1) chỉ còn \(x^2=\left\{1;9\right\}\) thỏa mãn \(\Rightarrow x=\left\{1;3\right\}\) \(\Rightarrow y=1\)
Vậy pt có 2 cặp nghiệm: \(\left(x;y\right)=\left(1;1\right);\left(3;1\right)\)
