Odpowiedź:
[tex]a)\\\\\frac{x^2-9}{2-x}\cdot\frac{x-2}{3+x}\\\\Zalo\.zenia\\\\2-x\neq 0\ \ \ \ i\ \ \ \ 3+x\neq 0\\\\-x\neq -2\ \ /\cdot(-1)\ \ \ \ i\ \ \ \ x\neq -3\\\\x\neq 2\ \ \ \ i\ \ \ \ x\neq -3\\\\D=R\setminus\left\{-3,2\right\}\\\\\\\frac{x^2-9}{2-x}\cdot\frac{x-2}{3+x}=\frac{(x-3)(x+3)}{-(x-2)}\cdot\frac{x-2}{3+x}=\frac{x-3}{-(x-2)}\cdot(x-2)=(x-3)\cdot(-1)=-x+3[/tex]
[tex]b)\\\\\frac{-x^2+3x}{x-1}:\frac{2x-6}{x^2-1}\\\\Zalo\.zenia\\\\x-1\neq 0\ \ \ \ i\ \ \ \ x^2-1\neq 0\ \ \ \ i\ \ \ \ 2x-6\neq 0\\\\x\neq 1\ \ \ \ i\ \ \ \ (x-1)(x+1)\neq 0\ \ \ \ i\ \ \ \ 2x\neq 6\ \ /:2\\\\x\neq 1\ \ \ \ i\ \ \ \ x\neq 1\ \ \ \ i\ \ \ \ x\neq -1\ \ \ \ x\neq 3\\\\D=R\setminus\left\{-1,1,3\right\}\\\\\\\frac{-x^2+3x}{x-1}:\frac{2x-6}{x^2-1}=\frac{-x^2+3x}{x-1}\cdot\frac{x^2-1}{2x-6}=\frac{-x(x-3)}{x-1}\cdot\frac{(x-1)(x+1)}{2(x-3)}=\frac{-x}{x-1}\cdot\frac{(x-1)(x+1)}{2}=[/tex]
[tex]=-x\cdot\frac{x+1}{2}=-\frac{x(x+1)}{2}=-\frac{x^2+x}{2}[/tex]
