c)
x^2-x-6\ne0 i x^2-3x\ne0 |zamieniam x^2-x-6=x^2-3x+2x-6=x(x-3)+2(x-3)=
(x-3)(x+2)\ne0 i x(x-3)\ne0
x\ne 3 , x\ne -2 , x\ne 0 , x\ne3
D \mathbb R \ {-2, 0, -3, 3}
\frac{x^2}{x^2-x-6} * \frac{x^2-9}{x^2-3x}=\frac{x^2}{x^2-x-6}*\frac{(x-3)(x+3)}{x(x-3)}=\frac{x(x+3)}{x^2-x-6}=\frac{x^2+3x}{x^2-x-6}
d)
\frac{x^{2}-25}{3x^{2}-2x} : \frac{x^{2}+10x+25}{2-3x}
x(3x-2)\ne0 i 2-3x\ne0
D = \mathbb R \ {0, 2/3, -5}
\frac{x^{2}-25}{3x^{2}-2x} : \frac{x^{2}+10x+25}{2-3x}=
\frac{x^{2}-25}{3x^{2}-2x} *\frac{2-3x}{x^{2}+10x+25}=
\frac{(x-5)(x+5)}{-x(2-3x)}*\frac{2-3x}{(x+5)^2}= |------ x\ne-5
-\frac{x-5}{x(x+5)}=\frac{5-x}{x^2+5x}