$(\frac{a^2-5a}{a^2-10a+25}+\frac{25}{a^2-25}):\frac{125-a^3}{5+a}=(\frac{a(a-5)}{(a-5{)}^2}+\frac{25}{(a-5)(a+5)}):\frac{125-a^3}{a+5}=(\frac{a}{a-5}+\frac{25}{(a-5)(a+5)})\ast \frac{a+5}{125-a^3}=\frac{a(a+5)+25}{(a-5)(a+5)}\ast \frac{a+5}{125-a^3}=\frac{a^2+5a+25}{a-5}\ast \frac{1}{(5-a)(25+5a+a^2)}=\frac{1}{a-5}\ast \frac{1}{5-a}=\frac{1}{a-5}\ast (-\frac{1}{a-5})=-\frac{1}{(a-5{)}^2}$
при a = 4,5:
$-\frac{1}{(4,5-5{)}^2}=-\frac{1}{(-0,5{)}^2}=-\frac{1}{(\frac{1}{2}{)}^2}=-\frac{1}{\frac{1}{4}}=-4$