[tex]x\neq1 \\ x\neq - 1 \\ \frac{x + 2}{x - 1} + \frac{x + 3}{x + 1} + \frac{x + 5}{1 - {x}^{2} } = 0 \\ \frac{x + 2}{x - 1} + \frac{x + 3}{x + 1} - \frac{x + 5}{(x - 1)(x + 1)} = 0 \\ \frac{(x + 2)(x + 1) + (x + 3)(x - 1) - (x + 5)}{(x - 1)(x + 1)} = 0 \\ {x}^{2} + x + 2x + 2 + {x}^{2} - x + 3x - 3 - x - 5 = 0 \\ 2 {x}^{2} + 4x - 6 = 0 \\ {x}^{2} + 2x - 3 = 0 \\ D = {2}^{2} - 4 \times ( - 3) = 4 + 12 = 16 \\ x_{1} = \frac{ - 2 - 4}{2} = - \frac{6}{2} = - 3 \\ x_{2} = \frac{ - 2 + 4}{2} = \frac{2}{2} = 1[/tex]

Второй корень не подходит, ответ: х = - 3