[1/(x-4)(x-3)] + [1/(x-3)(x-2)] + [ 1/(x-2)(x-1)] - [3/(x-4)(x-1)] +3 срочно нужно...

Решите задачу:

$1) \frac{1}{(x-4)(x-3)} + \frac{1}{(x-3)(x-2)}= \frac{x-2+x-4}{(x-4)(x-3)(x-2)} = \frac{2x-6}{(x-4)(x-3)(x-2)} = \\ = \frac{2(x-3)}{(x-4)(x-3)(x-2)} = \frac{2}{(x-4)(x-2)}$
$v2) \frac{2}{(x-4)(x-2)} + \frac{1}{(x-2)(x-1)} = \frac{2(x-1)+(x-4)}{(x-4)(x-2)(x-1)} = \frac{3x-6}{(x-4)(x-2)(x-1)} = \\= \frac{3}{(x-4)(x-1)} \\ 3) 0$