Question

Даю 85 баллов. У меня СОР

Answer 1

1. а)

$\frac{15 {x}^{2} }{6x + 15 {x}^{2} } = \frac{15 {x}^{2} }{3x(2 + 5x)} = \frac{5x}{2 + 5x}$

б)

$\frac{9x + 9y}{36 {x}^{2} - 36 {y}^{2} } = \frac{9(x + y)}{36( {x}^{2} - {y}^{2}) } = \frac{(x + y)}{4(x + y)(x - y)} = \frac{1}{4(x - y)}$

2. а)

$\frac{5}{12a - 12b} - \frac{3}{16a - 16b} = \frac{5}{12(a - b)} - \frac{3}{16(a - b)} = \frac{15 - 12}{48(a - b)} = \frac{3}{48(a - b)} = \frac{1}{16(a - b)}$

б)

$\frac{5x - 7}{ {x}^{2} - 4} + \frac{3x - 2}{x - 2} = \frac{5x - 7}{(x - 2)(x + 2)} + \frac{3x - 2}{x - 2} = \frac{5x - 7 + (3x - 2)(x + 2)}{(x - 2)(x + 2)} = \frac{5x - 7 + 3 {x}^{2} + 6x - 2x - 4 }{(x + 2)(x - 2)} = \frac{3 {x}^{2} + 9x - 11 }{(x + 2)(x - 2)}$

3.

$\frac{3x -9 {x}^{2} }{ {x}^{2} + 6x + 9} \div \frac{1 - 9 {x}^{2} }{ {x}^{2} - 9} = \frac{3x(1 - 3x)}{(x + 3)(x + 3)} \times \frac{(x + 3)(x - 3)}{(1 - 3x)(1 + 3x)} = \frac{3x}{(x + 3)} \times \frac{(x - 3)}{(1 + 3x)}$

теперь подставим

$\frac{3}{(1 + 3)} \times \frac{(1 - 3)}{(1 + 3)} = \frac{3}{4} \times \frac{( - 2)}{4} = \frac{3}{2} \times \frac{( - 1)}{4} = \frac{ - 3}{8}$