Question

Срочно !!!
Пожалуйста!!
Даю 20 баллов
Только с дополнительными множителями!!!

Answer 1

$\displaystyle \tt 1). \ \ \frac{8}{4-x}=-\frac{8}{x-4}=-\frac{8\cdot(x-4)}{(x-4)^{2}}=\frac{32-8x}{(x-4)^{2}};\\\\\\2). \ \ \frac{14b}{b^{2}-14b+49}=\frac{14b}{(b-7)^{2}}=\frac{28b}{2(b-7)^{2}}; \\\\\\{} \ \ \ \ \ \ \frac{b}{2b^{2}-14b}=\frac{b}{2b(b-7)}=\frac{1}{2(b-7)}=\frac{b-7}{2(b-7)^{2}};\\\\\\3). \ \ \frac{8+n}{n^{2}+6n+36}=\frac{(8+n)(n-6)}{(n^{2}+6n+36)(n-6)}=\frac{(8+n)(n-6)}{n^{3}-6^{3}};\\\\\\{} \ \ \ \ \ \ \frac{n}{n-6}=\frac{n(n^{2}+6n+36)}{(n^{2}+6n+36)(n-6)}=\frac{n(n^{2}+6n+36)}{n^{3}-6^{3}};$

$\displaystyle \tt 4). \ \ \frac{x}{x^{2}+2xy+y^{2}}=\frac{x}{(x+y)^{2}}=\frac{x(x-y)}{(x-y)(x+y)^{2}}; \\\\\\{} \ \ \ \ \ \ \frac{y}{x^{2}-y^{2}}=\frac{y}{(x-y)(x+y)}=\frac{y(x+y)}{(x-y)(x+y)^{2}};\\\\\\5). \ \ \frac{9+a}{a^{2}+9a+81}=\frac{(a+9)(a-9)}{(a^{2}+9a+81)(a-9)}=\frac{a^{2}-81}{a^{3}-9^{3}};\\\\\\{} \ \ \ \ \ \ \frac{2a}{a-9}=\frac{2a(a^{2}+9a+81)}{(a^{2}+9a+81)(a-9)}=\frac{2a(a^{2}+9a+81)}{a^{3}-9^{3}};$