Срочно помогите! Задание 2.4

Решите задачу:

$\frac{x + 2}{4x - 1} + \frac{x - 2}{4x + 1} = \frac{6x + 3}{16x {}^{2} - 1 } = 0 \\ \frac{x + 2}{4x - 1} + \frac{x - 2}{4x + 1} - \frac{6x + 3}{(4x - 1)(4x + 1)} = 0 \\ \frac{(4x + 1)(x + 2) + (4x - 1)(x - 2) - (6x + 3)}{(4x - 1)(4x + 1)} = 0 \\ \frac{4x {}^{2} + 8x + x + 2 + 4x {}^{2} - 8x - x + 2 - 6x - 3}{(4x - 1)(4x + 1)} = 0 \\ \frac{8x {}^{2} + 1 - 6x }{(4x - 1)(4x + 1)} = 0 \\ \frac{8x { }^{2} - 6x + 1}{(4x - 1)(4x + 1)} = 0 \\ \frac{8x {}^{2} - 2x - 6x + 1 }{(4x - 1)(4x + 1)} = 0 \\ \frac{2x(4x - 1) - (4x - 1)}{(4x - 1)(4x + 1)} = 0 \\ \frac{(4x - 1)(2x - 1)}{(4x - 1)(4x + 1)} = 0 \\ \frac{2x - 1}{4x + 1} = 0 \\ 2x - 1 = 0 \\ 2x = 1 \\ x = \frac{1}{2}$