\(\frac{16a^{5}b}{12a^{8}b^{2}} = \frac{4 ^{5} b^{1}}{3 ^{8} b^{2}} = \frac{4}{3a^{3}b}\)
\(\frac{ab + a^{2}}{a^{2}} = \frac{a(b + a)}{a^{2}} = \frac{b + a}{a}\)
\(\frac{x-3y}{x^{2}-9y^{2}} = \frac{x-3y}{(x-3y)(x+3y)} = \frac{1}{x+3y}\)
Ответ: а) \(\frac{4}{3a^{3}b}\); б) \(\frac{b + a}{a}\); в) \(\frac{1}{x+3y}\).