(3/25-a^2+1/a^2-10a+25)*(5-a)^2/a+3a/a+5=

Решите задачу:

$(\frac {3}{25-a^2}+\frac {1}{a^2-10a+25})*\frac{(5-a)^2}{a}+\frac {3a}{a+5}= (\frac {3*(5-a)}{(5-a)^2*(a+5)}+\frac {1*(a+5)}{(5-a)^2*(a+5)})*\frac{(5-a)^2}{a}+\frac {3a}{a+5}= (\frac {15-3a+a+5)}{(5-a)^2*(a+5)})*\frac{(5-a)^2}{a}+\frac {3a}{a+5}= (\frac {20-2a}{(5-a)^2*(a+5)})*\frac{(5-a)^2}{a}+\frac {3a}{a+5}= (\frac {20-2a}{(a+5)a})+\frac {3a}{a+5}=(\frac {20-2a+3a^2}{(a+5)a})$