9. (16 points) Given the functions
$$
f(x)=x+\frac{a}{x}-4, \quad g(x)=k x+3 .
$$

When $a \in[1,2]$, if the inequality
$$
\left|f\left(x_{1}\right)\right|-\left|f\left(x_{2}\right)\right|<g\left(x_{1}\right)-g\left(x_{2}\right)
$$

holds for any $x_{1}, x_{2} \in[2,4]\left(x_{1}<x_{2}\right)$, find the range of the real number $k$.
9. From the conditional inequality, we know that when
$$
\left|f\left(x_{1}\right)\right|-g\left(x_{1}\right)-1, \text{ it is sufficient that }
$$
$$
\sqrt{\frac{a}{k+1}} \geqslant 2+\sqrt{4-a} \text {, }
$$

i.e., $\square$
$$
\sqrt{\frac{1}{k+1}} \geqslant\left(\frac{2}{\sqrt{a}}+\sqrt{\frac{4}{a}-1}\right)_{\max }=2+\sqrt{3}
$$
$$
\Rightarrow -11 \text{ when } F(x)=(1-k) x+\frac{a}{x}-7 \text{ is monotonically decreasing on the interval } (2+\sqrt{4-a}, 4], \text{ } k>1 \text{ does not meet the requirement; }
$$
(3) When $k<1$, it is sufficient that
$$
\sqrt{\frac{a}{1-k}} \leqslant 2+\sqrt{4-a} \text {, }
$$

i.e., 
$$
\sqrt{\frac{1}{1-k}} \leqslant\left(\frac{2}{\sqrt{a}}+\sqrt{\frac{4}{a}-1}\right)_{\text {min }}=1+\sqrt{2}
$$
$$
\Rightarrow k \leqslant 2 \sqrt{2}-2 \text {. }
$$

In summary, $k \leqslant 6-4 \sqrt{3}$.