Question

2-$\frac{3}{x-2}=\frac{7}{x+2}$
$\frac{5}{x-1}+\frac{10}{x+1}=5$
Решите два примера пожалуйста.

Answer 1

1)
$2 - \frac{3}{x - 2} = \frac{7}{x + 2} \\ \frac{2(x - 2) - 3}{x - 2} - \frac{7}{x + 2} = 0 \\ \frac{2x -7 }{x - 2} - \frac{7}{x + 2} = 0 \\ \frac{(2x - 7)(x + 2) - 7(x - 2)}{(x - 2)(x + 2)} = 0 \\ \frac{2 {x}^{2} - 7x + 4x - 14 - 7x + 14}{(x - 2)(x + 2)} = 0 \\ \frac{2 {x}^{2} + 4x }{(x - 2)(x + 2)} = 0 \\ \frac{2x(x + 2)}{(x - 2)(x + 2)} = 0$
ОДЗ:
х ≠ +- 2

2х = 0, х = 0
х + 2 = 0, х = -2 -- не удовл.

Ответ: 0.

2)
$\frac{5}{x - 1} + \frac{10}{x + 1} = 5 \\ \frac{5(x + 1) + 10(x - 1) - 5(x + 1)(x - 1)}{(x - 1)(x + 1)} = 0 \\ \frac{5x + 5 + 10x - 10 - 5x {}^{2} + 5 }{(x - 1)(x + 1)} = 0 \\ \frac{ - 5 {x}^{2} + 15x }{(x - 1)(x + 1)} = 0 \\ \frac{ - 5x(x - 3)}{(x - 1)(x + 1)} = 0$
-5х = 0, х = 0
х - 3 = 0, х = 3

Ответ: 0; 3.