9. Given the inequality $1-l x \leq \frac{1}{\sqrt{x+1}} \leq 1-k x$ with the solution set $\left[0, \frac{1}{2}\right]$. Try to find the values of $l$ and $k$.

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9. Solution: It is easy to see that when $x=0$, the inequality obviously holds, so let $f(x)=\frac{1}{x}-\frac{1}{x \sqrt{x+1}}=\frac{\sqrt{x+1}-1}{x \sqrt{x+1}}$, then the original inequality is transformed into: $k \leq f(x) \leq l$ with the solution set $\left(0, \frac{1}{2}\right]$;

Let $t=\sqrt{x+1}$, then $f(t)=\frac{t-1}{\left(t^{2}-1\right) t}=\frac{1}{t(t+1)}, t \in\left(1, \frac{\sqrt{6}}{2}\right]$;
It is easy to see that $f(t)$ is monotonically decreasing on $(0,+\infty]$, and $f(t) \in\left[2-\frac{2 \sqrt{6}}{3}, \frac{1}{2}\right]$; hence $k \leq 2-\frac{2 \sqrt{6}}{3}, l \geq \frac{1}{2}$; since the solution set of $k \leq f(x) \leq l$ is $\left(0, \frac{1}{2}\right]$, it follows that $k=2-\frac{2 \sqrt{6}}{3}, l=\frac{1}{2}$.