Question

Решите уравнение методом замены переменной
$\frac{x-3}{6+2x} + \frac{4(6+2x)}{x-3} + 5 = 0$

Answer 1

Ответ:

замена:

$\frac{x - 3}{6 + 2x} = t \\ \\ t + \frac{4}{t} + 5 = 0 \: \: \: | \times t \\ {t}^{2} + 5t + 4 = 0 \\ D= 25 - 16 = 9 \\ t1 = \frac{ - 5 + 3}{2} = - 1 \\ t2 = - 4 \\ \\ \frac{x - 3}{6 + 2x} = - 1 \\ x - 3 = - (6 + 2x) \\ x - 3 = - 6 - 2x \\ 3x = - 3 \\ x_1 = - 1 \\ \\ \frac{x - 3}{6 + 2x} = - 4 \\ x - 3 = - 4(6 + 2x) \\ x - 3 = - 24 - 8x \\ 9x = - 21 \\ x _2 = - \frac{7}{3}$