Упростим выражение:
$$\frac{a+12}{4a+16} + \frac{a+4}{4a-16} + \frac{19}{a^2-16} = \frac{a+12}{4(a+4)} + \frac{a+4}{4(a-4)} + \frac{19}{(a-4)(a+4)} = \frac{(a+12)(a-4)}{4(a+4)(a-4)} + \frac{(a+4)(a+4)}{4(a-4)(a+4)} + \frac{19*4}{4(a-4)(a+4)} = \frac{a^2 + 12a - 4a - 48 + a^2 + 8a + 16 + 76}{4(a-4)(a+4)} = \frac{2a^2 + 16a + 44}{4(a-4)(a+4)} = \frac{2(a^2 + 8a + 22)}{4(a-4)(a+4)} = \frac{a^2 + 8a + 22}{2(a-4)(a+4)} = \frac{a^2 + 8a + 22}{2(a^2-16)} = \frac{a^2 + 8a + 22}{2a^2-32}$$
Ответ: $$\frac{a^2 + 8a + 22}{2a^2-32}$$