10. For any $x \in \mathbf{R}$, the function $f(x)$ satisfies the relation $f(x+2008)=f(x+2007)+f(x+2009)$, and $f(1)=\lg \frac{3}{2}, f(2)=\lg 15$, then the value of $f(2007)$ is $\qquad$ .
10. 1. By substituting $x-2007$ for $x$ in the given relation, we get $f(x)=f(x-1)+f(x+1)$.

Replacing $x$ with $x+1$, we get $f(x+1)=f(x)+f(x+2)$.
Adding (1) and (2) yields $f(x-1)+f(x+2)=0$.
Replacing $x$ with $x+1$, we get $f(x)+f(x+3)=0$, so $f(x+3)=-f(x), f(x+6)=f[(x+3)+3]=-f(x+3)=f(x)$. Therefore, $f(x)$ is a periodic function, and 6 is one of its periods. Thus, $f(2007)=f(6 \times 334+3)=f(3)$.
Also, $f(2)=f(1)+f(3), f(3)=f(2)-f(1)=\lg 15-\lg \frac{3}{2}=1$. Therefore, $f(2007)=f(3)=1$.
Note: The conclusion of this problem can be generalized as follows: If the function $y=f(x)(x \in \mathbf{R})$ satisfies $f(x)=f(x-a)+f(x+a)$, then the function $y=f(x)$ is a periodic function.
Proof: Since for any $x \in \mathbf{R}$, we have
$$
\begin{array}{l}
f(x)=f(x-a)+f(x+a), \\
f(x-a)=f(x-2 a)+f(x) .
\end{array}
$$
Adding (1) and (2) gives
$$
f(a+x)=-f(x-2 a), f(x)=-f(x-3 a) .
$$

Thus,
$$
f(x+6 a)=-f(x+6 a-3 a)=-f(x+3 a)=f(x+3 a-3 a)=f(x) \text {. }
$$

Therefore, the function $y=f(x)(x \in \mathbf{R})$ is a periodic function, and $6 a$ is one of its periods.