Ta có :
\(\dfrac{x^2+y^2}{5}=\dfrac{x^2-y^2}{3}\Leftrightarrow5\cdot\left(x^2-y^2\right)=3\cdot\left(x^2+y^2\right)\\ \Leftrightarrow5x^2-5y^2=3x^2+3y^2\\ \Leftrightarrow5x^2-3x^2=3y^2+5y^2\\ \Leftrightarrow2x^2=8y^2\\ \Leftrightarrow x^2=4y^2\)
Thay vào \(x^{10}\cdot y^{10}=1024,tacó:\)
\(x^{10}\cdot y^{10}=1024\Leftrightarrow\left(x^2\right)^5\cdot y^{10}=1024\\ \Leftrightarrow\left(4y^2\right)^5\cdot y^{10}=1024\\ \Leftrightarrow1024\cdot y^7\cdot y^{10}=1024\\ \Rightarrow y^{17}=1\\ \Rightarrow y=1\)
Mà \(x^2=4y^2\Rightarrow x^2=4\cdot1^2=4\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy \(\left(x;y\right)\in\left\{\left(2;1\right);\left(-2;1\right)\right\}\)
