(10) (15 points) Given the sequence of positive numbers $\left\{a_{n}\right\}(n \geqslant 0)$ satisfies $a_{n}=\frac{a_{n-1}}{m a_{n-2}}, n=2$,
$3, \cdots$, where $m$ is a real parameter. If $a_{2009}=\frac{a_{0}}{a_{1}}$, find the value of $m$.
10 Because $a_{2}=\frac{a_{1}}{m a_{0}}, a_{3}=\frac{\frac{a_{1}}{m a_{0}}}{m a_{1}}=\frac{1}{m^{2} a_{0}}, a_{4}=\frac{\frac{1}{m^{2} a_{0}}}{m \frac{a_{1}}{m a_{0}}}=\frac{1}{m^{2} a_{1}}$, $a_{5}=\frac{\frac{1}{m^{2} a_{1}}}{m \frac{1}{m^{2} a_{0}}}=\frac{a_{0}}{m a_{1}}, a_{6}=\frac{\frac{a_{0}}{m a_{1}}}{m \frac{1}{m^{2} a_{1}}}=a_{0}, a_{7}=\frac{a_{0}}{m \frac{a_{0}}{m a_{1}}}=a_{1}$, so
the sequence $\left\{a_{n}\right\}$ is a periodic sequence with a period of 6, therefore
$$
\frac{a_{0}}{a_{1}}=a_{2009}=a_{5}=\frac{a_{0}}{m a_{1}},
$$

thus we have $m=1$.