(a^3-15a^2+75a-125/a^2-10a+25)+(25/a+5) при a=1/2

Решите задачу:

$\frac{a^3-15a^2+75a-125}{a^2-10a+25}+\frac{25}{a+5}=\frac{a^3-5^3-15a(a-5)}{(a-5)^2}+\frac{25}{a+5}=\frac{(a-5)(a^2+5a+25)-15a(a-5)}{(a-5)^2}+\frac{25}{a+5}=\frac{(a-5)((a+5)^2-15a)}{(a-5)^2}+\frac{25}{a+5}=\frac{(a+5)^2-15a}{a-5}+\frac{25}{a+5}=\frac{(a+5)^3-15a(a+5)+25(a-5)}{(a-5)(a+5)}=\frac{a^3+15a^2+75a+125-15a^2-75a+25a-125}{(a-5)(a+5)}=\frac{a^3+25a}{(a-5)(a+5)}=\frac{a(a^2+25)}{a^2-25}=\frac{0,5(0,5^2+25)}{(0,5-5)(0,5+5)}=\frac{0,5(25,25)}{(-4,5)(5,5)}=\frac{12,625}{24,75}$