Question

помогите решить примеры пожалуйста,срочно!!!!!!!!!!!!

Answer 1

Ответ:

скачай фото матч.

помогает

Answer 2

$5)$

$\frac{x + 2}{x - 1} + \frac{x + 3}{x + 1} = \frac{x + 5}{ {x}^{2} - 1} \\ \frac{(x + 2)(x + 1) + (x + 3)(x - 1)}{( {x}^{} - 1)(x + 1)} = \frac{x + 5}{ {x}^{2} - 1 } \\ \frac{ {x}^{2} + x + 2x + 2 + {x}^{2} - x + 3x - 3 }{ {x}^{2} - 1} = \frac{x + 5}{ {x}^{2} - 1} \\ {2x}^{2} + 5x - 1 = x + 5 \\ {2x}^{2} - 4x - 6 = 0 \\ {x}^{2} - 2x -3 = 0 \\ x1 + x2 = - ( - 2) \\ x1 \times x2 = - 3 \\ \\ x1 + x2 = 2 \\ x1 \times x2 = - 3 \\ x1 = 3 \: \: \: \: \: x2 = - 1$

$6)$

$\frac{x - 1}{x + 3} + \frac{x + 1}{x - 3} = \frac{2x + 18}{ {x}^{2} - 9} \\ \frac{(x - 1)(x - 3) + (x + 1)(x + 3)}{(x + 3)(x - 3)} = \frac{2x + 18}{ {x}^{2} - 9} \\ \frac{ {x}^{2} - 3x - x + 3 + {x}^{2} + 3x + x + 3}{ {x}^{2} - 9} = \frac{2x + 18}{ {x}^{2} - 9 } \\ {2x}^{2} + 6 = 2x + 18 \\ {2x}^{2} - 2x - 12 = 0 \\ {x}^{2} - x - 6 = 0 \\ x1 + x2 = - ( - 1) \\ x1 \times x2 = - 6 \\ \\ x1 + x2 = 1 \\ x1 \times x2 = - 6 \\ x1 = 3 \: \: \: \: \: x2 = - 2$

$7)$

$\frac{x}{x - 3} + \frac{x - 3}{x + 3} = \frac{36}{ {x}^{2} - 9 } \\ \frac{x(x + 3) + (x - 3)(x - 3)}{(x - 3)(x + 3)} = \frac{36}{ {x}^{2} - 9 } \\ \frac{ {x}^{2} + 3x + {x}^{2} - 6x + 9 }{ {x}^{2} - 9} = \frac{36}{ {x}^{2} - 9 } \\ {2x}^{2} - 3x + 9 = 36 \\ {2x}^{2} - 3x - 27 = 0$

$D = {( - 3)}^{2} - 4 \times 2 \times ( - 27 )= 9 + 216 = 225$

$x1x2 = \frac{ - ( - 3)± \sqrt{225} }{2 \times 2} = \frac{3±15}{4} = - 3 \:; \: 4.5$