$ \frac{x-2}{x+5}>0 $ Since $\frac{a}{b}>0$, when $a>0$ and $b>0$ or $a<0$ and $b<0 b$ $\therefore$ either $x-2>0$ and $x+5>0$ or $x-2<0$ and $x+5<0$ Case $1: x-2>0$ and $x+5>0$ $\therefore x>2$ and $x>-5$ $\therefore \mathrm{x}>2$ $\therefore$ solution set $=(2, \infty)$ Case II: $ \begin{aligned} & x-2<0 \text { and } x+5<0 \\ & \therefore x<2 \text { and } x<-5 \\ & \therefore x<-5 \\ & \therefore \text { solution set }=(-\infty,-5) \end{aligned} $ $\therefore$ solution set $=(-\infty,-5)$ $\therefore$ the solution set of the given inequation is $(-\infty,-5) \cup(2, \infty)$
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