Выполните действия:
а) $\frac{a^2 - 25}{a + 3} * \frac{1}{a^2 + 5a} - \frac{a + 5}{a^2 - 3a}$;
б) $\frac{1 - 2x}{2x + 1} + \frac{x^2 + 3x}{4x^2 - 1} : \frac{3 + x}{4x + 2}$;
в) $\frac{b - c}{a + b} - \frac{ab - b^2}{a^2 - ac} * \frac{a^2 - c^2}{a^2 - b^2}$;
г) $\frac{a^2 - 4}{x^2 - 9} : \frac{a^2 - 2a}{xy + 3y} + \frac{2 - y}{x - 3}$.
$\frac{a^2 - 25}{a + 3} * \frac{1}{a^2 + 5a} - \frac{a + 5}{a^2 - 3a} = \frac{(a - 5)(a + 5)}{a(a + 3)(a + 5)} - \frac{a + 5}{a(a - 3)} = \frac{a - 5}{a(a + 3)} - \frac{a + 5}{a(a - 3)} = \frac{(a - 5)(a - 3) - (a + 5)(a + 3)}{a(a + 3)(a - 3)} = \frac{a^2 - 8a + 15 - (a^2 + 8a + 15)}{a(a + 3)(a - 3)} = -\frac{16a}{a(a + 3)(a - 3)} = -\frac{16}{a^2 - 9}$
$\frac{1 - 2x}{2x + 1} + \frac{x^2 + 3x}{4x^2 - 1} : \frac{3 + x}{4x + 2} = \frac{1 - 2x}{2x + 1} + \frac{x(x + 3)}{(2x - 1)(2x + 1)} * \frac{2(2x + 1)}{x + 3} = \frac{1 - 2x}{2x + 1} + \frac{2x}{2x - 1} = \frac{1 - 2x}{1 + 2x} - \frac{2x}{1 - 2x} = \frac{(1 - 2x)^2 - 2x(1 + 2x)}{(1 + 2x)(1 - 2x)} = \frac{1 - 4x + 4x^2 - 2x - 4x^2}{(1 + 2x)(1 - 2x)} = \frac{1 - 6x}{1 - 4x^2}$
$\frac{b - c}{a + b} - \frac{ab - b^2}{a^2 - ac} * \frac{a^2 - c^2}{a^2 - b^2} = \frac{b - c}{a + b} - \frac{b(a - b)}{a(a - c)} * \frac{(a - c)(a + c)}{(a - b)(a + b)} = \frac{b - c}{a + b} - \frac{b(a + c)}{a(a + b)} = \frac{a(b - c) - b(a + c)}{a(a + b)} = \frac{ab - ac - ab - bc}{a(a + b)} = -\frac{c(a + b)}{a(a + b)} = -\frac{c}{a}$
$\frac{a^2 - 4}{x^2 - 9} : \frac{a^2 - 2a}{xy + 3y} + \frac{2 - y}{x - 3} = \frac{(a - 2)(a + 2)}{(x - 3)(x + 3)} * \frac{y(x + 3)}{a(a - 2)} + \frac{2 - y}{x - 3} = \frac{y(a + 2)}{a(x - 3)} + \frac{2 - y}{x - 3} = \frac{y(a + 2) + a(2 - y)}{a(x - 3)} = \frac{ay + 2y + 2a - ay}{a(x - 3)} = \frac{2(a + y)}{a(x - 3)}$
Пожауйста, оцените решение
