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