Question

помогите пожалуйста ​

Answer 1

$2)\frac{2x+8c}{x^{2}-16c^{2}} =\frac{2(x+4c)}{(x+4c)(x-4c)}=\frac{2}{x-4c}\\\\4c-x=\frac{1}{4} \ \Rightarrow \ x-4c=-\frac{1}{4}\\\\\frac{2}{x-4c}=\frac{2}{-\frac{1}{4}}=2*(-4)= \boxed{-8} \\\\\\3)\frac{6}{a+b}=\frac{6*(a-b)}{(a+b)*(a-b)}=\frac{6(a-b)}{a^{2}-b^{2}} \\\\\frac{7}{a-b} =\frac{7*(a+b)}{(a-b)*(a+b)}=\frac{7(a+b)}{a^{2}-b^{2}} \\\\\\\frac{2b-c}{4b-2c}=\frac{2b-c}{2(2b-c)}=\frac{1}{2}=\frac{ab+2ac}{2(ab+2ac)}=\frac{b+2c}{2(b+2c)}$

$\frac{2a}{ab+2ac}=\frac{2a*2}{2(ab+2ac)}=\frac{4a}{2(ab+2ac)}=\frac{4}{2(b+2c)}$