ĐK: $x\geq \dfrac{-1}{8000}$
Đặt $\sqrt{1+8000x}=2y-1$, với $y\geq \dfrac{1}{2}$
Từ hệ ta có: $x^2-x=1000+1000(2y-1)$
$\Leftrightarrow x^2-x=2000y (1)$
Mặt khác: $\sqrt{1+8000x}=2y-1$
$\Leftrightarrow 4y^2-4y+1=1+8000x$
$\Leftrightarrow y^2-y=2000x (2)$
Từ $(1),(2)$ ta có hệ mới:
$\begin{cases}x^2-x=2000y\\y^2-y=2000x \end{cases}$
$\Leftrightarrow \left\{\begin{array}{l}x^2-x=2000y\\x^2-x-y^2+y=2000y-2000x\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}x^2-x=2000y\\(x-y)(x+y+1999)=0\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}x^2-x=2000y\\x=y\end{array}\right.$, vì $x+y+1999>0$
$\Leftrightarrow x=y=2001$, vì $y\geq \dfrac{1}{2}$.