Question

ПОМОГИТЕ пожалуйста срочно!!!!!!!!!!!!!!!!!!

Answer 1

$\frac{3a}{a-4}-\frac{a+2}{2a-8}*\frac{96}{a^2+2a}=\frac{3a}{a-4}-\frac{a+2}{2(a-4)}*\frac{2*48}{a(a+2)}=\frac{3a}{a-4}-\frac{48}{a(a-4)}=\\=\frac{3a^2-48}{a(a-4)}=\frac{3(a^2-16)}{a(a-4)}=\frac{3(a-4)(a+4)}{a(a-4)}=\frac{3a+12}{a}$