1) \(\frac{5-b}{3-a} - \frac{16-b^2}{a^2-6a+9} \cdot \frac{3-a}{b-4}\) 2) \(\frac{2-p}{3-m} + \frac{p^2-4}{m^2-9} \cdot \frac{m+3}{2-p}\)

Решение:

1. 1) \( \frac{5-b}{3-a} - \frac{16-b^2}{a^2-6a+9} \cdot \frac{3-a}{b-4} = \frac{5-b}{-(a-3)} - \frac{(4-b)(4+b)}{(a-3)^2} \cdot \frac{-(a-3)}{-(4-b)} = \frac{b-5}{a-3} - \frac{(4-b)(4+b)}{(a-3)^2} \cdot \frac{a-3}{4-b} = \frac{b-5}{a-3} - \frac{4+b}{a-3} = \frac{b-5 - (4+b)}{a-3} = \frac{b-5-4-b}{a-3} = \frac{-9}{a-3} \)
2. 2) \( \frac{2-p}{3-m} + \frac{p^2-4}{m^2-9} \cdot \frac{m+3}{2-p} = \frac{2-p}{-(m-3)} + \frac{(p-2)(p+2)}{(m-3)(m+3)} \cdot \frac{m+3}{-(p-2)} = \frac{p-2}{m-3} - \frac{(p-2)(p+2)}{(m-3)(m+3)} \cdot \frac{m+3}{p-2} = \frac{p-2}{m-3} - \frac{p+2}{m-3} = \frac{(p-2)-(p+2)}{m-3} = \frac{p-2-p-2}{m-3} = \frac{-4}{m-3} \)

Ответ: 1) \(\frac{-9}{a-3}\); 2) \(\frac{-4}{m-3}\).