Решите уравнение 3/1-х+1/х+1=28/1-х в квадрате

Решите задачу:

$\frac{3}{1-x} + \frac{1}{x+1} = \frac{28}{1-x^2} \\ \\ \frac{3}{1-x} + \frac{1}{1+x} = \frac{28}{(1-x)(1+x)} \\ \\ \frac{28}{(1-x)(1+x)} -\frac{3}{1-x} - \frac{1}{1+x} =0 \\ \\ \frac{28}{(1-x)(1+x)} -\frac{3+3x}{(1-x)(1+x)} - \frac{1-x}{(1-x)(1+x)} =0 \\ \\ \frac{28-3-3x-1+x}{(1-x)(1+x)} =0 \\ \\ \left \{ {{24-2x=0} \atop {x^2-1 \neq 0}} \left \{ {{x=12} \atop {x \neq б1}} \right. \\ \\ OTBET:12$