Question

Прошу Решить Как Можно Быстрее Задание По Алгебре 8 Класс. Заранее Благодарен

Answer 1

Ответ:

1.

а)

$\frac{5ab - 35b}{ {a}^{2} - 5 a - 14} \\ \\ {a}^{2} - 5 a - 14 =(a - 7)(a + 2) \\ D = 25 + 56 = 81 \\ a1 = \frac{5 + 9}{2} = 7 \\ a2 = - 2 \\ \\ \frac{5b(a - 7)}{(a - 7)(a + 2)} = \frac{5b}{a + 2}$

б)

$\frac{ {b}^{2} - 5b + 6}{5b - 15} \\ \\ {b}^{2} - 5 b + 6 = (b - 2)(b - 3) \\ D = 25 - 24 = 1 \\ b1 = \frac{5 + 1}{2} = 3 \\ b2 = 2 \\ \\ \frac{(b - 2)(b - 3)}{5(b - 3)} = \frac{b - 2}{5}$

2.

$( \frac{c + 2}{ {c}^{2} - c - 6} - \frac{c}{ {c}^{2} - 6c + 9} ) \times {(2c - 6)}^{2} \\ \\ {c}^{2} - c - 6 =(c - 3)(c + 2) \\ D = 1 + 24 = 25 \\ c1 = \frac{1 + 5}{2} = 3 \\ c2 = - 2 \\ \\ {c}^{2} - 6c + 9 = {(c - 3)}^{2} \\ \\ ( \frac{c + 2}{( c - 3)(c + 2)} - \frac{c}{ {(c - 3)}^{2} } ) \times {(2c - 6)}^{2} = \\ = (\frac{1}{c - 3} - \frac{c}{ {(c - 3)}^{2} } ) \times {(2(c - 3))}^{2} = \\ = \frac{c - 3 - c}{ {(c - 3)}^{2} } \times 4 {(c - 3)}^{2} = \\ = - 3 \times 4 = - 12$