(\(3x^2\left(y^2-10\right)+\left(y^2-10\right)=507\)y^2-10)=507
\(\left(y^2-10\right)\left(3x^2+1\right)=507=3.169=3.13.13\)
VP chia hết cho 3 và \(\left\{{}\begin{matrix}3x^2+1\ge1\\\left(3x^2+1\right)=3k+1\end{matrix}\right.\)=>
Loại hệ nghiệm âm, (y^2-10)=3k
\(\left[{}\begin{matrix}y^2-10=3=13\left(loai\right)\\y^2-10=3.13=49\end{matrix}\right.\) \(\left\{{}\begin{matrix}y^2-10=3.13\\3x^2+1=13\end{matrix}\right.\) \(\left\{{}\begin{matrix}y^2=49\\3x^2=12\end{matrix}\right.\)\(\Rightarrow3x^2y^2=12\cdot49=588\)