9. Given the set $A=\left\{x \left\lvert\, \frac{x+8}{x-5} \leqslant 0\right.\right\}$, and the non-empty set $B=\{x \mid t+1 \leqslant x \leqslant 2 t-1\}$. If $A \cap B=\varnothing$, then the range of real number $t$ is $\qquad$ .
9. $t \geqslant 4$.

From the given, $A=\{x \mid-8 \leqslant x<5\}$, since $A \cap B=\varnothing$, and $B \neq \varnothing$, we have $\left\{\begin{array}{l}2 t-1 \geqslant t+1 \\ 2 t-1<-8\end{array}\right.$ or $\left\{\begin{array}{l}2 t-1 \geqslant t+1 \\ t+1 \geqslant 5,\end{array}\right.$ which gives $t \geqslant 4$.