1. Given that $a$ and $b$ are the roots of the equation $\log _{3 x} 3+\log _{77}(3 x)=$ $-\frac{4}{3}$. Then $a+b=(\quad)$.
(A) $\frac{10}{27}$
(B) $\frac{4}{81}$
(C) $\frac{10}{81}$
(D) $\frac{28}{81}$
$-1 . C$.
The original equation is transformed into
$$
\frac{1}{1+\log _{3} x}+\frac{1+\log _{3} x}{3}=-\frac{4}{3} \text {. }
$$

Let $1+\log _{3} x=t$, then $\frac{1}{t}+\frac{t}{3}=-\frac{4}{3}$.
Solving for $t$ yields $t_{1}=-1, t_{2}=-3$.
Therefore, $1+\log _{3} x=-1$ or $1+\log _{3} x=-3$.
Thus, the two roots of the equation are $\frac{1}{9}$ and $\frac{1}{81}$, and their sum is $\frac{10}{81}$.