a)
\frac{2x-6}{x^3-9x}+\frac{x^2}{x^3(x+3)}
dziedzina
x^3-9x\ne0
x(x^2-9)\ne0
x(x-3)9x+3)\ne0
x\ne 0 , x-3\ne0 , x+3\ne 0
x\ne 0 , x\ne 3 , x\ne -3
i
x^3(x+3)\ne0
x\ne 0 , x\ne -3
D=\mathbb R - \{-3,0,3\}
{\frac{2x-6}{x^3-9x}+\frac{x^2}{x^3(x+3)}=\frac{2(x-3)}{x(x^2-9)}+\frac{x^2}{x^3(x+3)}=\frac{2(x-3)}{x(x-3)(x+3)}+\frac{x^2}{x^3(x+3)}=}
{=\frac{2}{x(x+3)}+\frac{x^2}{x^3(x+3)}=\frac{2x^2+x^2}{x^3(x+3)}=\frac{3x^2}{x^3(x+3)}=\frac{3}{x(x+3)} }
b)
\frac{4x^2+4}{4x+4}:\frac{x^2-2x+1}{5x^2-5}
dziedzina
4x+4\ne0
4(x+1)\ne0
x+1\ne 0
x\ne -1
i
5x^2-5\ne 0
5(x^2-1)\ne 0 \ |:5
5(x-1)(x+1)\ne 0
x-1\ne 0 i x+1\ne0
x\ne 1 , x\ne -1
D=\mathbb R - \{-1,1\}
{\frac{4x^2+4}{4x+4}:\frac{x^2-2x+1}{5x^2-5}=\frac{\not4(x^2+1)}{\not4(x+1)}\cdot \frac{5x^2-5}{x^2-2x+1}=\frac{x^2+1}{x+1}\cdot \frac{5(x^2-1)}{(x-1)^2}=\frac{5(x^2+1)\cdot (x-1)(x+1)}{(x+1)\cdot (x-1)(x-1)}=\frac{5(x^2+1)}{x-1}}