Question

b-1/(b+3)^2-b/(b^2-9) помогите пожалуйста

Answer 1

$\dfrac{b-1}{(b+3)^2}-\dfrac{b}{b^2-9}=\dfrac{b-1}{(b+3)^2}-\dfrac{b}{(b-3)(b+3)}=\dfrac{(b-1)(b-3)-b(b+3)}{(b+3)^2(b-3)}=\\ \\ \\ =\dfrac{b^2-4b+3-b^2-3b}{(b+3)^2(b-3)}=\dfrac{3-7b}{(b+3)^2(b-3)}$