$\frac{2x - 1}{x + 4} - \frac{3x + 1}{4 - x} = \frac{4x^2 + 8}{x^2 - 16}$
$\frac{2x - 1}{x + 4} + \frac{3x + 1}{x - 4} = \frac{4x^2 + 8}{(x - 4)(x + 4)}$
x + 4 ≠ 0
x ≠ −4
и
x − 4 ≠ 0
x ≠ 4
$\frac{(2x - 1)(x - 4) + (3x + 1)(x + 4)}{(x - 4)(x + 4)} = \frac{4x^2 + 8}{(x - 4)(x + 4)} |* (x - 4)(x + 4)$
$2x^2 - x - 8x + 4 + 3x^2 + x + 12x + 4 = 4x^2 + 8$
$5x^2 + 4x + 8 - 4x^2 - 8 = 0$
$x^2 + 4x = 0$
x(x + 4) = 0
x = 0
или
x + 4 = 0
x = −4 − не подходит
Ответ: Г) 0