\(\left\{{}\begin{matrix}\dfrac{3}{x^2}=2x+y\\\dfrac{3}{y^2}=2y+x\end{matrix}\right.\left(đk:x;y\ne0\right)\Leftrightarrow\left\{{}\begin{matrix}2x^3+x^2y=3\\2y^3+xy^2=3\end{matrix}\right.\left(1\right)\\ \Leftrightarrow2x^3+x^2y=2y^3+xy^2\\ \Leftrightarrow\left(2x^3-2y^3\right)+\left(x^2y-xy^2\right)=0\\ \Leftrightarrow\left(2x^3-2y^3\right)+\left(x^2y-xy^2\right)=0\\ \Leftrightarrow2\left(x-y\right)\left(x^2+xy+y^2\right)+xy\left(x-y\right)=0\\ \Leftrightarrow\left(x-y\right)\left(2x^2+2xy+2y^2+xy\right)=0\\ \Leftrightarrow\left(x-y\right)\left(2x^2+3xy+2y^2\right)=0\\ \Leftrightarrow x-y=0\left(Vì\text{ }2x^2+3xy+2y^2\ne0\right)\\ \Leftrightarrow x=y\left(2\right)\)
Thay \(\left(2\right)\) vào \(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}2x^3+x^3=3\\2y^3+y^3=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x^3=3\\3y^3=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
Vậy................