Question

‼️‼️Допоможіть зробити даю 40 балів будь ласка срочно ‼️‼️

Answer 1

Ответ:

$\frac{22 + 11a}{45 - 9a} = \frac{11(2 + a)}{9(5 - a)}$

$\frac{1}{x + 6} - \frac{x}{x - 4} = \frac{x - 4 - {x}^{2} - 6}{(x + 6)(x - 4)} = \frac{ - {x}^{2} + x - 10 }{ {x}^{2} - 2x - 24 }$

$\frac{ {p}^{2} - 81}{3p + 27} = \frac{(p - 9)(p + 9)}{3(p + 9)} = \frac{p - 9}{3} = \frac{p}{3} - 3$

$\frac{ {a}^{2} - 49 }{ {a}^{2} + 14a + 49 } = \frac{(a - 7)(a + 7)}{ {(a + 7)}^{2} } = \frac{a - 7}{a + 7}$

$\frac{8b - 40c}{3b - 15c} = \frac{8(b - 5c)}{3(b - 5c)} = \frac{8}{3} = 2 \frac{2}{3}$

$\frac{x - y}{3y - 3x} = \frac{x - y}{3(y - x)} = - \frac{1}{3}$

$\frac{4}{a - 2} + \frac{3a + 2}{ {a}^{2} - 2a} = \frac{4 }{a - 2} + \frac{3a + 2}{a(a - 2)} = \frac{4a + 3a + 2}{a(a - 2)} = \frac{7a + 2}{a(a - 2)}$

$\frac{a + 5}{a - 5} + \frac{20a}{25 - {a}^{2} } = \frac{a + 5}{a - 5} + \frac{20a}{(5 - a)(5 + a)} = \frac{ - {(a + 5)}^{2} + 20a }{25 - {a}^{2} } = \frac{ - {a}^{2} - 10a - 25 + 20a}{ 25 - {a}^{2}} = \frac{ - {a}^{2} + 10a - 25 }{25 - {a}^{2} } = \frac{ {(a - 5)}^{2} }{(a - 5)(5 + a)} = \frac{a - 5}{5 + a}$

$\frac{6a}{a - 7} - \frac{21}{a + 7} + \frac{6 {a}^{2} }{49 - {a}^{2} } = \frac{6a(a + 7) - 21(a - 7) - 6 {a}^{2} }{(a - 7)(a + 7)} = \frac{21(9 - a)}{ {a}^{2} - 49}$

$\frac{3b - 1}{6b + 2} - \frac{6b}{1 - 9 {b}^{2} } - \frac{3b + 1}{9b - 3} = \\ = \frac{3b - 1}{2(3b + 1)} + \frac{6b}{(3b - 1)(1 + 3b)} - \frac{3b + 1}{3(3b - 1)} = \\ = \frac{3 {(3b - 1)}^{2} + 36b - 2 {(3b + 1)}^{2} }{6(3b - 1)(3b + 1)} = \\ = \frac{27{b}^{2} - 18b + 3 + 36b - 18 {b}^{2} - 12b - 2}{6(3b - 1)(3b + 1)} = \\ = \frac{9 {b}^{2} + 6b + 1 }{6(3b - 1)(3b + 1)} = \\ = \frac{3b + 1}{6(3b - 1)}$