The real values of $x$ satisfying the inequality: ${\log }_{0.5}(x^2-25)-{\log }_{0.5}(x-5)<{\log }_{0.5}3$ is:

Step 1: DomainArguments of logs must be positive:$x^2-25>0⟹(x-5)(x+5)>0⟹x>5$ or $x<-5$.$x-5>0⟹x>5$.Intersection: $x>5$.Step 2: Solve inequality
$${\log }_{0.5}\left(\frac{x^2-25}{x-5}\right)<{\log }_{0.5}3$$
$${\log }_{0.5}(x+5)<{\log }_{0.5}3$$
Since the base $0.5<1$, the inequality reverses when removing the logs:
$$x+5>3$$
$$x>-2$$
Combining with the domain $x>5$, we get $x>5$.