Помогите пожалуйста с 3 и 4))))

№3.
$\frac{a^2-1}{a^2} - \frac{a^2-9}{a} * \frac{1}{a+3} = \frac{a^2-1}{a^2} - \frac{(a-3)(a+3)}{a} * \frac{1}{a+3} = \frac{a^2-1}{a^2} - \frac{a-3}{a} =$
$= \frac{a^2-1-a(a-3)}{a^2} = \frac{a^2-1-a^2+3a}{a^2} = \frac{-1+3a}{a^2}$
№4.
$(\frac{x-y}{x+y} - \frac{x+y}{x-y} ): \frac{4}{x^2-y^2} = \frac{(x-y)(x-y)-(x+y)(x+y)}{(x+y)(x-y)} : \frac{4}{x^2-y^2} = \frac{(x-y)^2-(x+y)^2}{(x+y)(x-y)} :$
$: \frac{4}{x^2-y^2} = \frac{(x-y-(x+y))(x-y+x+y)}{(x+y)(x-y)} : \frac{4}{x^2y^2} = \frac{(x-y-x-y)*2x}{(x+y)(x-y)} : \frac{4}{x^2-y^2} =$
$\frac{(-2y)*2x}{(x+y)(x-y)} : \frac{4}{x^2y^2} = \frac{-4xy}{(x+y)(x-y)} : \frac{4}{x^2-y^2} = \frac{-4xy}{(x+y)(x-y)} * \frac{x^2-y^2}{4} =$
$=- \frac{xy}{(x+y)(x-y)} *(x-y)(x+y)=-xy$