Đặt: $y=1-2014x^2$, khi đó ta có hệ:
$\left\{\begin{array}{l}y=1-2014x^2\\x=1-2014y^2\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}y=1-2014x^2\\x-y=2014x^2-2014y^2\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}y=1-2014x^2\\(x-y)(2014x+2014y+1)=0\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}x=y\\y=1-2014x^2\end{array}\right.\\\left\{\begin{array}{l}2014x+2014y+1=0\\y=1-2014x^2\end{array}\right.\end{array}\right.$
Đến đây dễ dàng giải ra nghiệm: $x\in\{\dfrac{1\pm\sqrt{8053}}{4028};\dfrac{-1\pm\sqrt{8057}}{4028}\}$