Question

Даю 21 балл помогите срочно

Answer 1

1) х-4≠0
х≠4
Ответ: при х≠4

2)
$1) \frac{24 {a}^{6} {b}^{4} }{ {16}^{} {a}^{3} {b}^{7} } = \frac{3 {a}^{3} }{2 {b}^{3} }$
$2) \frac{15x - 10xy}{5xy} = \frac{5x(3 - 2y)}{5xy} = \frac{3 - 2y}{y}$
$3) \frac{ {m}^{ 2} - 4}{2m - 4} = \frac{(m - 2)(m + 2)}{2(m - 2)} = \frac{m + 2}{2}$
$4) \frac{25 - {a}^{2} }{ {a}^{2} - 10a + 25 } = \frac{(5 - a)(5 + a)}{ {(a - 5)}^{2} } = \frac{5 + a}{a - 5}$
3)
$1) \frac{x - 8}{4 {x}^{2} } - \frac{5 - 12x}{6 {x}^{3} } = \frac{3x(x - 8) - 2(5 - 12x)}{12 {x}^{3} } = \frac{3 {x}^{2} - 24x - 10 + 24x }{12 {x}^{3} } = \frac{3 {x}^{2} - 10}{12 {x}^{3} }$
$2) \frac{20}{ {a}^{2} + 4a} - \frac{5}{a} = \frac{20}{a(a + 4)} - \frac{5}{a} = \frac{20 - 5(a + 4)}{a(a + 4)} = \frac{20 - 5a - 20}{a(a + 4)} = \frac{ - 5a}{a(a + 4)} = - \frac{5}{a + 4}$
$3)\frac{ {m}^{2} }{ {m}^{2} - 9} - \frac{m}{m + 3} = \frac{ {m}^{2} }{(m - 3)(m + 3)} - \frac{m}{m + 3} = \frac{ {m}^{2} - m(m - 3) }{(m - 3)(m + 3)} = \frac{ {m}^{2} - {m}^{2} + 3m }{(m - 3)(m + 3)} = \frac{3m}{(m - 3)(m + 3)}$
$4)2p - \frac{14 {p}^{2} }{7p + 3} = \frac{2p( 7p + 3)- 14 {p}^{2} }{7p + 3} = \frac{14 {p}^{2} + 6p - 14 {p}^{2} }{7p + 3} = \frac{6p}{7p + 3}$
4)
$1)\frac{y + 3}{2y + 2} - \frac{y + 1}{2y - 2} + \frac{3}{ {y}^{2} - 1 } = \frac{y + 3}{2(y + 1)} - \frac{y + 1}{2(y - 1)} + \frac{3}{(y - 1)(y + 1)} = \frac{(y + 3)(y - 1) - (y + 1)(y + 1) + 3 \times 2}{2(y - 1)(y + 1)} = \frac{ {y}^{2} - y + 3y - 3 - {y}^{2} - y - y - 1 + 6}{2(y - 1)(y + 1)} = \frac{2}{2(y - 1)(y + 1)} = \frac{1}{(y - 1)(y + 1)}$
$2) \frac{2 {b}^{2} - b }{ {b}^{3} + 1 } - \frac{b - 1}{ {b}^{2} - b + 1 } = \frac{b(2b - 1)}{(b + 1)({b}^{2} - b + 1)} - \frac{b - 1}{ {b}^{2} - b + 1 } = \frac{b(2b - 1) - (b - 1)(b + 1)}{(b + 1)({b}^{2} - b + 1)} = \frac{2 {b}^{2} - b - {b}^{2} + 1}{(b + 1)({b}^{2} - b + 1)} = \frac{ {b}^{2} - b + 1 }{(b + 1)({b}^{2} - b + 1)} = \frac{1}{(b + 1)}$