Regarding the equation $2007 x^{3}+2006 x^{2}+2005 x=0$, it is correct to state:
(a) It has no roots
(b) It has 3 distinct real roots
(c) It has 2 equal roots
(d) It has only one real root
(e) It has 3 positive roots
Let's observe that $x=0$ is a solution, so options (a) and (e) are discarded. Now we only need to study the solutions of $2007 x^{2}+2006 x+2005=0$, which is a second-degree equation with discriminant

$$
\begin{aligned}
\Delta & =2006^{2}-4 \times 2007 \times 2005=2006^{2}-4(2006+1)(2006-1) \\
& =2006^{2}-4\left(2006^{2}-1\right)=-3 \times 2006^{2}+4<0
\end{aligned}
$$

thus it has no real roots. Therefore, the initial equation has only one real root, and the correct option is (d).

Note: Another way (and simpler) to show that $\Delta<0$ is: since $2006<2007$ and $2006<4 \times 2005$, then

$$
2006 \times 2006<4 \times 2005 \times 2007 \Longrightarrow 2006^{2}-4 \times 2005 \times 2007<0
$$