Example 6 Let the function $f(x)=\frac{(k+1) x^{2}+(k+3) x+(2 k-8)}{(2 k-1) x^{2}+(k+1) x+(k-4)}$ have its domain denoted by $D$, then what is the set of real numbers $k$ that makes $f(x)>0$ for all $x \in D$?
If $k+1=0$, i.e., $k=-1$, then $f(x)=\frac{10-2 x}{3 x^{2}+5}$, it is easy to see that $D=\mathbf{R}$. But when $x>5$, $f(x)0$ for $x \in D$ always holds, then $x_{1}$ and $x_{2}$ must be the roots of $(2 k-1) x^{2}+(k+1) x+(k-4)=0$. Therefore, we have
$$
\left\{\begin{array}{l}
\frac{k+1}{2 k-1}>0, \\
x_{1}+x_{2}=-\frac{k+3}{k+1}=-\frac{k+1}{2 k-1}, \Rightarrow k=1 . \\
x_{1} x_{2}=\frac{2 k-8}{k+1}=\frac{k-4}{2 k-1}, \\
\Delta=(k+3)^{2}-4(k+1)(2 k-8) \geqslant 0,
\end{array}\right.
$$

In fact, when $k=1$, $f(x)=2>0$, where $D=(-\infty,-3) \cup(-3,1) \cup(1,+\infty)$.
If $(k+1) x^{2}+(k+3) x+(2 k-8)=0$ has no real roots, then the equivalent condition of the problem is
$$
\begin{array}{l}
\left\{\begin{array}{l}
\frac{k+1}{2 k-1}>0, \\
\Delta_{1}=(k+3)^{2}-4(k+1)(2 k-8)\frac{1}{2} \text { or } k\frac{15+16 \sqrt{2}}{7} \text { or } k\frac{15+16 \sqrt{2}}{7} \text { or } k\frac{15+16 \sqrt{2}}{7}$ or $\left.k<\frac{15-16 \sqrt{2}}{7}\right\}$.
\end{array}\right.
$$