9. 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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The function $f(x)=|\lg (x+1)|$ is defined, and real numbers $a, b (a<b)$ satisfy the conditions $f(a)=f\left(-\frac{b+1}{b+2}\right)$ and $f(10 a+6 b+21)=4 \lg 2$. Determine the values of $a$ and $b$.
$$
f(a)=f\left(-\frac{b+1}{b+2}\right) \Rightarrow|\lg (a+1)|=\left|\lg \left(1-\frac{b+1}{b+2}\right)\right|=\left|\lg \frac{1}{b+2}\right|
$$
$\Rightarrow a+1=b+2$ or $a+1=\frac{1}{b+2}$, since $a<b$, then $a+1=\frac{1}{b+2}$.
$f(10 a+6 b+21)=4 \lg 2 \Rightarrow 10 a+6 b+22=16$ or $10 a+6 b+22=\frac{1}{16}$.
When $10 a+6 b+22=16$,
$$
\begin{array}{l}
\frac{10}{b+2}+6 b=4 \Rightarrow \frac{5}{b+2}+3 b=2 \Rightarrow 5+(b+2)(3 b-2)=0 \\
\Rightarrow 3 b^{2}+4 b+1=0 \Rightarrow b=-1 \text { or }-\frac{1}{3} .
\end{array}
$$

If $b=-1 \Rightarrow a=0$, which does not meet the condition; if $b=-\frac{1}{3} \Rightarrow a+1=\frac{3}{5} \Rightarrow a=-\frac{2}{5}$. When $10 a+6 b+22=\frac{1}{16}$,
$$
\frac{10}{b+2}+6 b=\frac{1}{16}-12=-\frac{191}{16} \Rightarrow 160+(b+2)(96 b+191)=0
$$
$\Rightarrow 96 b^{2}+383 b+382=0 \Rightarrow \Delta=383^{2}-4 \times 96 \times 382<0$ no solution.
In summary, we have $a=-\frac{2}{5}, b=-\frac{1}{3}$.