Question

Помогите решить примеры по алгебре

Answer 1

Ответ:

за последнее извини, не поняла цифры

Answer 2

$3) \: \\ a) \frac{a + 3}{a - 1} - \frac{a}{1 - a} = \frac{a + 3}{a - 1} + \frac{a}{a - 1} = \frac{a + 3 + a}{a - 1} = \frac{2a + 3}{a - 1} \\ b) \frac{3x + 2y}{2x - 3y} - \frac{x - 8y}{3y - 2x} = \frac{3x + 2y}{2x - 3y} + \frac{x - 8y}{2x - 3y} = \frac{3x + 2y + x - 8y}{2x - 3y} = \frac{4x - 6y}{2x - 3y} = \frac{2(2x - 3y)}{2x - 3y} = 2 \\ c) \: \frac{ {b}^{2} }{2b - 10} + \frac{25}{10 - 2b} = \frac{ {b}^{2} }{2b - 10} - \frac{25}{2b - 10} = \frac{ {b}^{2} - 25 }{2b - 10} = \frac{(b - 5)(b + 5)}{2(b - 5)} = \frac{b + 5}{2}$

$\frac{9y + 1}{ {y}^{2} - 4} - \frac{y - 8}{4 - {y}^{2} } + \frac{1 - 7y}{ {y}^{2} - 4 } = \frac{9y + 1}{ {y}^{2} - 4} + \frac{y - 8}{ {y}^{2} - 4 } + \frac{1 - 7y}{ {y}^{2} - 4 } = \frac{9y + 1 + y - 8 + 1 - 7y}{ {y}^{2} - 4} = \frac{3y - 6}{ {y}^{2} - 4 } = \frac{3(y - 2)}{(y - 2)(y + 2)} = \frac{3}{y + 2}$

$\frac{3x}{ {x}^{3} - 1 } - \frac{4x - 1}{1 - {x}^{3} } - \frac{ {x}^{3} }{1 - {x}^{3} } = \frac{3x}{ {x}^{3} - 1 } + \frac{4x - 1}{ {x}^{3} - 1} + \frac{ {x}^{3} }{ {x}^{3} - 1} = \frac{3x + 4x - 1 + {x}^{3} }{ {x}^{3} - 1} = \frac{7x - 1 + {x}^{3} }{ {x}^{3} - 1 }$