9. (16 points) Let the function $f(x)=|\lg (x+1)|$, and real numbers $a, b (a<b)$ satisfy
$$
f(a)=f\left(-\frac{b+1}{b+2}\right), f(10 a+6 b+21)=4 \lg 2 \text {. }
$$

Find the values of $a, b$.
From the given, we have
$$
\begin{array}{l}
|\lg (a+1)|=\left|\lg \left(-\frac{b+1}{b+2}+1\right)\right| \\
=\left|\lg \frac{1}{b+2}\right|=|\lg (b+2)| .
\end{array}
$$

Thus, $a+1=b+2$ or $(a+1)(b+2)=1$.
Given $a1 \text {. }$
$$

Therefore, $f(10 a+6 b+21)$
$$
=\lg \left[6(b+2)+\frac{10}{b+2}\right]=4 \lg 2 \text {. }
$$

Hence, $6(b+2)+\frac{10}{b+2}=16$.
Solving this, we get $b=-\frac{1}{3}$ or $b=-1$ (discard).
Substituting $b=-\frac{1}{3}$ into equation (1), we get $a=-\frac{2}{5}$.
Therefore, $a=-\frac{2}{5}, b=-\frac{1}{3}$.