Выполните действия:
1) $\frac{2}{x} + \frac{3x - 2}{x + 1}$;
2) $\frac{m}{n} - \frac{m}{m + n}$;
3) $\frac{a}{a - 3} - \frac{3}{a + 3}$;
4) $\frac{c}{3c - 1} - \frac{c}{3c + 1}$;
5) $\frac{x}{2y + 1} - \frac{x}{3y - 2}$;
6) $\frac{a - b}{b} - \frac{a - b}{a + b}$.
$\frac{2}{x} + \frac{3x - 2}{x + 1} = \frac{2(x + 1) + x(3x - 2)}{x(x + 1)} = \frac{2x + 2 + 3x^2 - 2x}{x(x + 1)} = \frac{3x^2 + 2}{x(x + 1)}$
$\frac{m}{n} - \frac{m}{m + n} = \frac{m(m + n) - mn}{n(m + n)} = \frac{m^2 + mn - mn}{n(m + n)} = \frac{m^2}{n(m + n)}$
$\frac{a}{a - 3} - \frac{3}{a + 3} = \frac{a(a + 3) - 3(a - 3)}{(a - 3)(a + 3)} = \frac{a^2 + 3a - 3a + 9}{(a - 3)(a + 3)} = \frac{a^2 + 9}{a^2 - 9}$
$\frac{c}{3c - 1} - \frac{c}{3c + 1} = \frac{c(3c + 1) - c(3c - 1)}{(3c - 1)(3c + 1)} = \frac{3c^2 + c - 3c^2 + c}{(3c - 1)(3c + 1)} = \frac{2c}{9c^2 - 1}$
$\frac{x}{2y + 1} - \frac{x}{3y - 2} = \frac{x(3y - 2) - x(2y + 1)}{(2y + 1)(3y - 2)} = \frac{3xy - 2x - 2xy - x}{(2y + 1)(3y - 2)} = \frac{xy - 3x}{(2y + 1)(3y - 2)}$
$\frac{a - b}{b} - \frac{a - b}{a + b} = \frac{(a - b)(a + b) - b(a - b)}{b(a + b)} = \frac{a^2 - b^2 - ab + b^2}{b(a + b)} = \frac{a^2 - ab}{b(a + b)}$
Пожауйста, оцените решение