Question

прошу помогите !!!Очень надо
даю 20 баллов

Answer 1

$\frac{x - 2}{x + 1} + \frac{x + 1}{x - 2} = 4 \frac{1}{4} \\ \frac{x - 2}{x + 1} + \frac{x + 1}{x - 2} = \frac{17}{4}$
х ≠ -1, х ≠ 2
$t = \frac{x - 2}{x + 1} \\ t + \frac{1}{t} = \frac{17}{4} \\ t + \frac{1}{t} - \frac{17}{4} = 0 \\ \frac{4 {t}^{2} + 4 - 17t }{4t} = 0 \\ 4 {t}^{2} - 17t + 4 = 0$
Решаем с помощью дискриминанта:
$d = {b}^{2} - 4ac = 289 - 4 \times 4 \times 4 = 289 - 64 = 225 \\ \sqrt{d} = 15$
Находим t:
$t1 = \frac{ - b + \sqrt{d} }{2a} = \frac{17 + 15}{8} = 4 \\ t2 = \frac{ - b - \sqrt{d} }{2a} = \frac{17 - 15}{8} = \frac{2}{8} = \frac{1}{4}$
Подставляем:
$t = \frac{x - 2}{x + 1}$
↓
$\frac{x - 2}{x + 1} = 4\\ x - 2 = 4(x + 1) \\ x - 2 = 4x + 4 \\ - 3x = 6 \\ x1 = - 2 \\ \\ \frac{x - 2}{x + 1} = \frac{1}{4} \\ (x - 2) \times 4 = x + 1 \\ 4x - 8 = x + 1 \\ 3x = 9 \\ x2 = 3$
Ответ: х1 = -2, х2 = 3