Упростите выражение:
1) $\frac{5b}{b - 3} - \frac{b + 6}{2b - 6} * \frac{90}{b^2 + 6b}$;
2) $\frac{b + 2}{b^2 - 2b + 1} : \frac{b^2 - 4}{3b - 3} - \frac{3}{b - 2}$.
$\frac{5b}{b - 3} - \frac{b + 6}{2b - 6} * \frac{90}{b^2 + 6b} = \frac{5b}{b - 3} - \frac{b + 6}{2(b - 3)} * \frac{90}{b(b + 6)} = \frac{5b}{b - 3} - \frac{1}{b - 3} * \frac{45}{b} = \frac{5b}{b - 3} - \frac{45}{b(b - 3)} = \frac{5b^2 - 45}{b(b - 3)} = \frac{5(b^2 - 9)}{b(b - 3)} = \frac{5(b - 3)(b + 3)}{b(b - 3)} = \frac{5(b + 3)}{b}$
$\frac{b + 2}{b^2 - 2b + 1} : \frac{b^2 - 4}{3b - 3} - \frac{3}{b - 2} = \frac{b + 2}{(b - 1)^2} : \frac{(b - 2)(b + 2)}{3(b - 1)} - \frac{3}{b - 2} = \frac{b + 2}{(b - 1)^2} * \frac{3(b - 1)}{(b - 2)(b + 2)} - \frac{3}{b - 2} = \frac{1}{b - 1} * \frac{3}{b - 2} - \frac{3}{b - 2} = \frac{3}{b - 2}(\frac{1}{b - 1} - 1) = \frac{3}{b - 2} * \frac{1 - (b - 1)}{b - 1} = \frac{3}{b - 2} * \frac{1 - b + 1}{b - 1} = \frac{3}{b - 2} * \frac{2 - b}{b - 1} = -\frac{3}{2 - b} * \frac{2 - b}{b - 1} = -\frac{3}{b - 1} = \frac{3}{1 - b}$
Пожауйста, оцените решение
