Помогите срочно. Зарание спасибо

Question 1

Question 2

В1.
$\frac{a {}^{3} }{a {}^{2} - 4 } + \frac{a}{2 - a} - \frac{2}{a + 2} - a + 4 = \\ \\ = \frac{ {a}^{3} }{(a - 2)(a + 2)} - \frac{a}{a - 2} - \frac{2}{a + 2} - \frac{a {}^{3} - 4a}{ {a}^{2} - 4 } + \frac{4 {a}^{2} - 16}{ {a}^{2} - 4 } = \\ \\ = \frac{ {a}^{3} }{ {a}^{2} - 4} - \frac{a {}^{2} + 2a }{a {}^{2} - 4} - \frac{2a - 4}{ {a}^{2} - 4} - \frac{a {}^{3} - 4a}{ {a}^{2} - 4} + \frac{4 {a}^{2} - 16 }{ {a}^{2} - 4} = \\ \\ = \frac{ {a}^{3} - {a}^{2} - 2a - 2a + 4 - {a}^{3} + 4a + 4 {a}^{2} - 16 }{ {a}^{2} - 4} = \\ \\ = \frac{3 {a}^{2} - 12 }{ {a}^{2} - 4 } = \\ \\ = \frac{3(a {}^{2} - 4)}{ {a}^{2} - 4} = \\ \\ = \frac{3(a - 2)(a + 2)}{(a - 2)(a + 2)} = \\ \\ = 3$
Ответ: 3

С1.
$\frac{1}{x(x + 1)} + \frac{1}{(x + 1)(x + 2)} + \frac{1}{(x + 2)(x + 3)} - \frac{3}{x(x + 3)} - 3 = \\ \\ = \frac{x {}^{2} + 5x + 6 + x {}^{2} + 3x + x {}^{2} + x - 3x {}^{2} - 9x - 6 - 3(x {}^{2} + x)(x {}^{2} + 5x + 6)}{x(x + 1)(x + 2)(x + 3)} = \\ \\ = \frac{ - 3(x {}^{4} + 5x {}^{3} + 6x {}^{2} + x {}^{3} + 5x {}^{2} + 6x)}{x(x + 1)(x + 2)(x + 3)} = \\ \\ = \frac{ - 3x {}^{4} - 15x {}^{3} - 18x {}^{2} - 3x {}^{3} - 15x {}^{2} - 18x }{x {}^{4} + 6x {}^{3} + 11x {}^{2} + 6x } = \\ \\ = \frac{ - 3x {}^{4} - 18x {}^{3} - 33x {}^{2} - 18x}{x {}^{4} + 6x {}^{3} + 11x {}^{2} + 6x } = \\ \\ = \frac{ - x(3x {}^{3} + 18x {}^{2} + 33x + 18) }{x(x {}^{3} + 6x {}^{2} + 11x + 6) } = \\ \\ = \frac{ - (3x {}^{3} + 33x)(18x {}^{2} + 18) }{(x {}^{3} + 11x)(6x {}^{2} + 6) } = \\ \\ = - \frac{3x(x {}^{2} + 11) + 18(x {}^{2} + 1) }{x(x {}^{2} + 11) + 6(x {}^{2} + 1) } = \\ \\ = - \frac{3x + 18}{x + 6} = \\ \\ = - \frac{3(x + 6)}{x + 6} = \\ \\ = - 3$
Ответ: -3