\(\dfrac{y^2-x^2}{3}=\dfrac{x^2+y^2}{5}\Leftrightarrow5y^2-5x^2=3x^2+3y^2\\ \Leftrightarrow2y^2-8x^2=0\\ \Leftrightarrow2\left(y^2-4x^2\right)=0\\ \Leftrightarrow2\left(y-2x\right)\left(y+2x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}y=2x\\y=-2x\end{matrix}\right.\)
Vì \(x^{10};y^{10}>0\) nên 2 trường hợp có nghiệm tương đương
Với \(y=2x\Leftrightarrow x^{10}\cdot y^{10}=x^{10}\cdot2^{10}\cdot x^{10}=1024\)
\(\Leftrightarrow x^{20}\cdot1024=1024\Leftrightarrow x^{20}=1\Leftrightarrow\left[{}\begin{matrix}x=1\Rightarrow y=2\\x=-1\Rightarrow y=-2\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left\{\left(1;2\right);\left(-1;-2\right)\right\}\)
