Question

помогите пожалуйста ​

Answer 1

Ответ:

$\dfrac{b^{2}+15}{b^{2}+5b};$

$\dfrac{4z}{z^{2}-4};$

Объяснение:

$\dfrac{3}{b}+\dfrac{b-3}{b+5}=\dfrac{3(b+5)+b(b-3)}{b(b+5)}=\dfrac{3b+15+b^{2}-3b}{b(b+5)}=\dfrac{b^{2}+15}{b^{2}+5b};$

$\dfrac{2z^{2}}{z^{2}-4}-\dfrac{2z}{z+2}=\dfrac{2z^{2}}{z^{2}-2^{2}}-\dfrac{2z}{z+2}=\dfrac{2z^{2}}{(z-2)(z+2)}-\dfrac{2z}{z+2}=\dfrac{2z^{2}-2z(z-2)}{(z-2)(z+2)}=$

$=\dfrac{2z^{2}-2z^{2}+4z}{(z-2)(z+2)}=\dfrac{4z}{z^{2}-4};$