\frac{x^2-2x+1}{(x-1)(x+1)}
(x-1)(x+1)\ne 0
x-1\ne 0 i x+1\ne0
x\ne 1 i x\ne -1
D=\mathbb R \backslash \{-1, 1\}
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\frac{x^2-2x+1}{(x-1)(x+1)}=\frac{(x-1)^2}{(x-1)(x+1)}=\frac{(x-1)(x-1)}{(x-1)(x+1)}=\frac{x-1}{x+1}
dla x=-2
\frac{x-1}{x+1}=\frac{-2-1}{-2+1}=\frac{-3}{-1}=3