12. Given real numbers $a, b$ satisfy
$$
6^{a}=2010,335^{b}=2010 \text {. }
$$

Then the value of $\frac{1}{a}+\frac{1}{b}$ is $\qquad$
12. 1 .

Notice, $6^{a b}=2010^{b}, 335^{a b}=2010^{a}$.
Then $(6 \times 335)^{a b}=2010^{a+b}$.
Thus $a b=a+b \Rightarrow \frac{1}{a}+\frac{1}{b}=1$.