Срочно!! докажите тождество

Решите задачу:

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