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

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9 Because $f(a)=f\left(-\frac{b+1}{b+2}\right)$, so
$$
\begin{aligned}
|\lg (a+1)| & =\left|\lg \left(-\frac{b+1}{b+2}+1\right)\right| \\
& =\left|\lg \left(\frac{1}{b+2}\right)\right|=|\lg (b+2)|,
\end{aligned}
$$

Therefore, $a+1=b+2$ or $(a+1)(b+2)=1$, and since $a1 .
\end{aligned}
$$

Thus,
$$
\begin{aligned}
f(10 a+6 b+21) & =\left|\lg \left[6(b+2)+\frac{10}{b+2}\right]\right| \\
& =\lg \left[6(b+2)+\frac{10}{b+2}\right] .
\end{aligned}
$$

Also,
$$
f(10 a+6 b+21)=4 \lg 2,
$$

so
$$
\lg \left[6(b+2)+\frac{10}{b+2}\right]=4 \lg 2,
$$

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 $(a+1)(b+2)=1$ yields $a=-\frac{2}{5}$.
Therefore, $a=-\frac{2}{5}, b=-\frac{1}{3}$.