Question

помогите пожалуйста!​

Answer 1

Ответ:

Номер 1

$\\ \frac{12a {}^{3} x - 16a {}^{2}x {}^{2} }{20ax {}^{3} - 15a {}^{2} x {}^{2} } = \frac{ax \times (12a {}^{2} - 16ax) }{ax \times (20x {}^{2} - 15ax) } = \frac{ - 4a \times ( - 3a + 4x)}{x \times( 20x - 15a)} = \frac{12a {}^{2} - 16ax }{x \times 5(4x - 3a)} = \frac{ - 4a \times ( - 3a + 4x)}{x \times 5(4x - 3a)} = \frac{ - 4a}{x \times 5} = - \frac{ 4a}{5x} \\ \\ \frac{x {}^{2} - 6x + 9 }{9x {}^{2} - 27x} = \frac{(x - 3 ) {}^{2} }{9x \times (x - 3)} = \frac{x - 3}{9x}$

----------------------------

Номер 2

$\\ \frac{2}{5x} + \frac{5}{9x} = \frac{9 \times 2}{9 \times 5x} + \frac{5 \times 5}{5 \times 9x} = \frac{18}{45x} + \frac{25}{45x} = \frac{18 + 25}{45x} = \frac{43}{45x} \\ \\ \frac{1}{x - 4} - \frac{1}{x + 4} = \frac{x + 4 - (x - 4)}{(x - 4) \times (x + 4)} = \frac{x + 4 - x + 4}{x {}^{2} - 16 } = \frac{8}{x {}^{2} - 16 } \\ \\ 13a {}^{2} \times \frac{5b}{26a {}^{3} } = a {}^{2} \times \frac{5b}{2a {}^{3} } = \frac{5b}{2a} \\ \\ \frac{15x {}^{2} y}{7a {}^{3} } \div \frac{18y}{35a {}^{2} b} = \frac{15x {}^{2}y }{7a {}^{3} } \times \frac{35a {}^{2}b }{18y} = \frac{5x {}^{2} }{a} \times \frac{5b}{6 } = \frac{25bx {}^{2} }{6a}$

--------------------------

Номер 3

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