\[\frac{(b - 2)^{2} + 8b + 1}{b} = \frac{b^{2} - 4b + 4 + 8b + 1}{b} =\]
\[= \frac{b^{2} + 4b + 5}{b} =\]
\[= \frac{b^{2}}{b} + \frac{4b}{b} + \frac{5}{b} = b + 4 + \frac{5}{b} \in Z\ \ \]
\[при\ \ b = 1;b = - 1;b = 5;b = - 5.\]
\[\mathbf{\ }\frac{22p^{4}q²}{99p^{5}q} = \frac{2q}{9p}\]