Упростите выражение:
1) $\frac{5n - 1}{20n} - \frac{7n - 8}{20n} - \frac{8n + 7}{20n}$;
2) $\frac{9m + 2}{m^2 - 4} - \frac{m - 9}{4 - m^2} + \frac{1 - 7m}{m^2 - 4}$;
3) $\frac{3k}{k^3 - 1} + \frac{4k + 1}{1 - k^3} + \frac{k^2}{1 - k^3}$.
$\frac{5n - 1}{20n} - \frac{7n - 8}{20n} - \frac{8n + 7}{20n} = \frac{5n - 1 - (7n - 8) - (8n + 7)}{20n} = \frac{5n - 1 - 7n + 8 - 8n - 7}{20n} = \frac{-10n}{20n} = -\frac{1}{2}$
$\frac{9m + 2}{m^2 - 4} - \frac{m - 9}{4 - m^2} + \frac{1 - 7m}{m^2 - 4} = \frac{9m + 2}{m^2 - 4} + \frac{m - 9}{m^2 - 4} + \frac{1 - 7m}{m^2 - 4} = \frac{9m + 2 + m - 9 + 1 - 7m}{m^2 - 4} = \frac{3m - 6}{m^2 - 4} = \frac{3(m - 2)}{(m - 2)(m + 2)} = \frac{3}{m + 2}$
$\frac{3k}{k^3 - 1} + \frac{4k + 1}{1 - k^3} + \frac{k^2}{1 - k^3} = \frac{3k}{k^3 - 1} - \frac{4k + 1}{k^3 - 1} - \frac{k^2}{k^3 - 1} = \frac{3k - (4k + 1) - k^2}{k^3 - 1} = \frac{3k - 4k - 1 - k^2}{k^3 - 1} = \frac{-k^2 - k - 1}{(k - 1)(k^2 + k + 1)} = \frac{-(k^2 + k + 1)}{(k - 1)(k^2 + k + 1)} = \frac{-1}{k - 1} = -\frac{1}{k - 1} = \frac{1}{1 - k}$
Пожауйста, оцените решение
