Срочно. решите пожалуйста алгебру.

Решите задачу:

$\frac{10}{ {a}^{2} - 4} - \frac{3}{a - 2} + \frac{4}{a + 2} = \frac{10}{(a - 2)(a + 2)} - \frac{3(a + 2)}{(a - 2)(a + 2)} + \frac{4(a - 2)}{(a - 2)(a + 2)} = \frac{10 - (3a + 6) + 4a - 8}{(a - 2)(a + 2) } = \frac{10 - 3a - 6 + 4a - 8}{(a - 2)(a + 2)} = \frac{a - 4}{(a - 2)(a + 2)} = \frac{a - 4}{ {a}^{2} - 4 }$

$( \frac{a}{a - b} - \frac{a}{a + b} ) \times \frac{a - b}{ab} = ( \frac{a(a + b)}{(a - b)(a + b)} - \frac{a(a - b)}{(a - b)(a + b)} ) \times \frac{a - b}{ab} = \frac{ {a}^{2} + ab - {a}^{2} + ab }{(a - b)(a + b)} \times \frac{a - b}{ab} = \frac{2ab \times (a - b)}{(a - b)(a + b) \times ab} = \frac{2}{a + b}$

$\frac{2c}{c - 3} - \frac{ {c}^{2} + c}{4} \div \frac{c + 1}{8} = \frac{2c}{c - 3} - \frac{ {c}^{2} + c }{4} \times \frac{8}{c + 1} = \frac{2c}{c - 3} - \frac{c(c + 1) \times 8}{4 \times (c + 1)} = \frac{2c}{c - 3} - \frac{2c}{1} = \frac{2c - 2c(c - 3)}{c - 3} = \frac{2c - 2 {c}^{2} + 6c}{c - 3} = \frac{ - 2 {c}^{2} + 8c }{c - 3}$