Помогите плиз сделать задания

№2.
$( \frac{x}{x-3} - \frac{2}{x+3}) : \frac{4x^2+4x +24}{x^2-9} = \\ \\ = \frac{x(x+3) -2(x-3) }{(x-3)(x+3)} * \frac{x^2-9}{4(x^2+x+6)} = \\ \\ = \frac{x^2 +3x -2x + 6}{x^2 - 3^2} * \frac{x^2 - 9}{4(x^2 +x+ 6)} = \\ \\ = \frac{x^2 +x + 6}{x^2 - 9 } * \frac{x^2 - 9 }{ 4(x^2 +x+6)} = \\ \\ = \frac{1*1}{1*4} = \frac{1}{4} =0.25$

№3.
$\frac{9-p^2}{3p+9} * \frac{p^2}{(3-p)^2} + \frac{p}{p-3} = - \frac{p}{3} \\ \\ \frac{(3-p)(3+p)}{3(3+p)} * \frac{p^2}{(3-p)(3-p)} + \frac{p}{p-3} = - \frac{p}{3} \\ \\ \frac{p^2 }{3*(3-p)} - \frac{p}{3-p} = - \frac{p}{3} \\ \\ \frac{p^2 - 3p}{3(3-p)} = - \frac{p}{3} \\ \\ \frac{-p(-p + 3)}{3(3-p)} = - \frac{p}{3} \\ \\ -\frac{p(3-p)}{3(3-p)} = - \frac{p}{3} \\ \\ - \frac{p}{3} = - \frac{p}{3}$