2. Simplify the expression $\frac{1+9 a^{-1}+20 a^{-2}}{1+8 a^{-1}+16 a^{-2}} \cdot\left(a^{2}+4 a\right) \cdot\left(1-25 a^{-2}\right)^{-1}$.
2. A power with a negative integer exponent in the first factor can be written in the form of a fraction. In the numerator and denominator of the double fraction, we find a common denominator. In the numerator, we get $\frac{a^{2}+9 a+20}{a^{2}}$, and in the denominator, $\frac{a^{2}+8 a+16}{a^{2}}$. We eliminate the double fraction, factor the numerator into $(a+$ 4) $(a+5)$ and the denominator into $(a+4)^{2}$. We factor the second factor into $a(a+4)$. We simplify the last (third) factor $\left(1-\frac{25}{a^{2}}\right)^{-1}=\frac{a^{2}}{a^{2}-25}$ and factor the denominator into $\frac{a^{2}}{(a+5)(a-5)}$. We simplify the fractions and get $\frac{a^{3}}{a-5}$.

Writing powers with fractions $\frac{1+\frac{9}{a}+\frac{20}{a^{2}}}{1+\frac{8}{a}+\frac{16}{a^{2}}}$.

1 point

Arrangement of the first factor $\frac{\frac{a^{2}+9 a+20}{a^{2}}}{\frac{a^{2}+8 a+16}{a^{2}}}$. . 1 point

Factoring the numerator and denominator $\frac{(a+4)(a+5)}{(a+4)^{2}}$. . . 1 point
Factoring the second factor $a(a+4)$ ..... 1 point
Arrangement of the third factor $\left(1-\frac{25}{a^{2}}\right)^{-1}=\frac{a^{2}}{a^{2}-25}=\frac{a^{2}}{(a+5)(a-5)}$ ..... 1 point
Result $\frac{a^{3}}{a-5}$ ..... 1 point