Question

помогите по алгебре пж дам 100б​

Answer 1

Ответ:

$\frac{c^{2}-9 }{2c+8}*\frac{4c+16}{c^{2}+6c+9 }=\frac{(c+3)(c-3)}{2c+8} *\frac{2(2c+8)}{(c+3)^{2} }=\frac{(c-3)2}{c+3}=\frac{2(1.8-3)}{1.8+3}=\frac{-2.4}{4.8} =-2$