Выполните действия: ам,
(с-0 (1-е)
а+2 а

с) +—

2а-а` а’-4

ь

5 2х__ _е ь
2х—6 х-3х

а Зт+п 4т

Эт? —Зтп ти?

Выполните действия: ам,(с-0 (1-е)а+2 ас) +—2а-а` а’-4ь5 2х__ _е ь2х—6 х-3ха Зт+п 4тЭт? —Зтп ти?

Ответ:

megabobr11

18.06.2021 08:04

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

$\frac{2x}{2x - 6} - \frac{9}{ {x}^{2} - 3x } = \frac{2x}{2(x - 3)} - \frac{9}{x(x - 3)} = \\ = \frac{2x \times x - 9 \times 2}{2x(x - 3)} = \frac{2 {x}^{2} - 18}{2x(x - 3)} = \frac{2(x - 3)(x + 3)}{2x(x - 3)} = \\ = \frac{x + 3}{x}$

$\frac{a + 2}{2a - a {}^{2} } + \frac{a}{a {}^{2} - 4} = \frac{a + 2}{ - a(a - 2)} + \frac{a}{( a - 2)(a + 2)} = \\ = \frac{(a + 2)(a + 2) - {a}^{2} }{ - a(a - 2)(a + 2)} = - \frac{ {a}^{2} + 4a + 4 - a {}^{2} }{ a(a - 2)(a + 2)} = \\ = - \frac{4(a + 1)}{a(a - 2)(a + 2)}$

$\frac{3m + n}{9 {m}^{2} - 3 mn} - \frac{4n}{9m {}^{2} - n {}^{2} } = \\ = \frac{3m + n}{3m(3m - n)} - \frac{4n}{(3m - n)(3m + n)} = \\ = \frac{(3m + n)(3m + n) - 4n \times 3m}{3m(3m - n)(3m + n)} = \\ = \frac{9 {m}^{2} + 6mn + n {}^{2} - 12mn }{3m(3m - n)(3m + n)} = \frac{9 {m}^{2} - 6mn + n {}^{2} }{3m(3m - n)(3m + n)} = \\ = \frac{(3m - n) {}^{2} }{3m(3m - n)(3m + n)} = \frac{3m - n}{3m(3m + n)}$