\frac{9x^2-6x+1}{(3x+1)(3x-1)}
3x+1\ne 0 i 3x-1\ne 0
3x\ne -1 i 3x\ne1
x\ne -\frac{1}{3} i x\ne \frac{1}{3}
D=\mathbb R\backslash \{-\frac{1}{3},\frac{1}{3}\}
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\frac{9x^2-6x+1}{(3x+1)(3x-1)}=\frac{(3x-1)^2}{(3x+1)(3x-1)}=\frac{(3x-1)(3x-1)}{(3x+1)(3x-1)}=\frac{3x-1}{3x+1}
dla x=-1
\frac{3x-1}{3x+1}=\frac{3*(-1)-1}{3*(-1)+1}=\frac{-3-1}{-3+1}=\frac{-4}{-2}=2
wzór skróconego mnożenia
a^2-2ab+b^2=(a-b)^2