\[\frac{16-b^2}{b^2-b-12}\]
\[16-b^2 = (4-b)(4+b)\]
\[b^2-b-12=0\]
\[D = (-1)^2 - 4 \cdot 1 \cdot (-12) = 1 + 48 = 49\]
\[b_1 = \frac{1 + \sqrt{49}}{2 \cdot 1} = \frac{1 + 7}{2} = \frac{8}{2} = 4\]
\[b_2 = \frac{1 - \sqrt{49}}{2 \cdot 1} = \frac{1 - 7}{2} = \frac{-6}{2} = -3\]
\[b^2-b-12 = (b-4)(b+3)\]
\[\frac{16-b^2}{b^2-b-12} = \frac{(4-b)(4+b)}{(b-4)(b+3)} = -\frac{(b-4)(4+b)}{(b-4)(b+3)} = -\frac{4+b}{b+3}\]
Ответ: \(-\frac{4+b}{b+3}\)