Ответ: ↓

[tex] \frac{b + 2}{ {b}^{2} - 2b + 1} \div \frac{ {b}^{2} - 4}{3b - 3} - \frac{3}{b - 2} = \frac{3}{1 - b} \\ \frac{b + 2}{ {(b - 1)}^{2} } \div \frac{(b - 2)(b + 2)}{3(b - 1)} - \frac{3}{b - 2} = \frac{3}{1 - b} \\ \frac{b + 2}{ {(b - 1)}^{2}} \times \frac{3(b - 1)}{(b - 2)(b + 2)} - \frac{3}{b - 2} = \frac{3}{1 - b} \\ \frac{3}{(b - 1)(b - 2)} - \frac{3}{b - 2} = \frac{3}{b - 1} \\ \frac{3 - 3(b - 1)}{(1 - b)(b - 2)} = \frac{3}{1 - b} \\ \frac{3 - 3b + 3}{b - 2b - {b}^{2} + 2b } = \frac{3}{1 -b} \\ \frac{ - 3b}{b - {b}^{2} } = \frac{3}{1 - b} [/tex]

[tex] - 3b + {3b}^{2} = 3b - 3 {b}^{2} \\ - 3b + {3b}^{2} - 3b + {3b}^{2} = 0 \\ {6b}^{2} - 6b = 0 \\ 6b(b - 1) = 0 \\ \\ 6b = 0 \: \: \: \: \: b = 0 \\ b - 1 = 0 \: \: \: \: \: b = 1[/tex]