Odpowiedź:
[tex]a)\ \ \frac{x}{6-x}:\frac{x^2+x}{x^2-36}\\\\6-x\neq 0\ \ \ \ i\ \ x^2-36\neq 0\ \ \ \ i\ \ \ \ x^2+x\neq 0\\\\\\6-x\neq 0\\\\-x\neq -6\ \ /\cdot(-1)\\\\x\neq 6\\\\\\x^2-36\neq 0\\\\(x-6)(x+6)\neq 0\\\\x-6\neq 0\ \ \ \ i\ \ \ \ x+6\neq 0\\\\x\neq 6\ \ \ \ \ \ i\ \ \ \ x\neq -6\\\\\\x^2+x\neq 0\\\\x(x+1)\neq 0\\\\x\neq 0\ \ \ \ i\ \ \ \ x+1\neq 0\\\\x\neq 0\ \ \ \ i\ \ \ \ x\neq -1\\\\D=R\setminus\left\{-6,-1,0,6\right\}[/tex]
[tex]\frac{x}{6-x}:\frac{x^2+x}{x^2-36}=\frac{x}{-(x-6)}\cdot\frac{x^2-36}{x^2+x}=\frac{x}{-(x-6)}\cdot\frac{(x-6)(x+6)}{x(x+1)}=\frac{1}{-(x-6)}\cdot\frac{(x-6)(x+6)}{x+1}=\\\\=-\frac{1}{1}\cdot\frac{x+6}{x+1}=-\frac{x+6}{x+1}[/tex]