Question

(2 supra x-1 minus 1supra x+1): (x+3)(×-1) supra ×^2-2x+1=?

Answer 1

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

Answer 2

aducem I paranteza la acelasi numitor care este (x+1)(x-1)
 rezulta {[2(x+1)-(x-1)]/(x+1)(x-1}·(x-1)²/(x+3)(x-1)= simplifica   x-1
[(2x+1-x+1)/(x+1)(x-1)]·(x-1)/(x+3)=simplificam cu x-1
(x+2)/(x+1)(x+3)
cond.x≠-1 si x≠-3
x+2=0
x=-2