Упр.179 ГДЗ Мерзляк Полонский 8 класс (Алгебра)

1) (x2+14x+49)/(x+6) : (13/(x+6) — x + 6);
2) (c — (2c-9)/(c+8) : (c2+3c)/(c2-64) + 24/c; Выполните действия:
1) (b+4)/(b2-6b+9) : (b2-16)/(2b-6) — 2/(b-4);
2) ((m-1)/(m+1) — (m+1)/(m-1)) : 4m/(m2-1);

1. $$\frac{x^2+14x+49}{x+6}:\left(\frac{13}{x+6}-x+6\right)=\frac{(x+7)^2}{x+6}:\frac{13-(x-6)(x+6)}{x+6}$$
   $$=\frac{(x+7)^2}{x+6}:\frac{49-x^2}{x+6}=\frac{(x+7)^2}{x+6}\cdot\frac{x+6}{(7-x)(7+x)}$$
   $$=\frac{x+7}{7-x}.$$

2. $$\left(c-\frac{2c-9}{c+8}\right):\frac{c^2+3c}{c^2-64}+\frac{24}{c}$$
   $$=\frac{c(c+8)-(2c-9)}{c+8}\cdot\frac{(c-8)(c+8)}{c(c+3)}+\frac{24}{c}$$
   $$=\frac{c^2+6c+9}{1}\cdot\frac{c-8}{c(c+3)}+\frac{24}{c}$$
   $$=\frac{(c+3)^2(c-8)}{c(c+3)}+\frac{24}{c}=\frac{(c+3)(c-8)}{c}+\frac{24}{c}$$
   $$=\frac{c^2-5c}{c}=c-5.$$

3. $$\left(\frac{b+4}{b^2-6b+9}:\frac{b^2-16}{2b-6}\right)-\frac{2}{b-4}$$
   $$=\frac{b+4}{(b-3)^2}\cdot\frac{2(b-3)}{(b-4)(b+4)}-\frac{2}{b-4}$$
   $$=\frac{2}{(b-3)(b-4)}-\frac{2}{b-4}$$
   $$=\frac{2-2(b-3)}{(b-3)(b-4)}=\frac{8-2b}{(b-3)(b-4)}$$
   $$=\frac{-2(b-4)}{(b-3)(b-4)}=\frac{2}{3-b}.$$

4. $$\left(\frac{m-1}{m+1}-\frac{m+1}{m-1}\right):\frac{4m}{m^2-1}$$
   $$=\frac{(m-1)^2-(m+1)^2}{(m+1)(m-1)}:\frac{4m}{(m-1)(m+1)}$$
   $$=\frac{m^2-2m+1-m^2-2m-1}{(m+1)(m-1)}:\frac{4m}{(m-1)(m+1)}$$
   $$=\frac{-4m}{(m-1)(m+1)}\cdot\frac{(m-1)(m+1)}{4m}=-1.$$

Ответ

1. $$\frac{x+7}{7-x}$$
2. $$c-5$$
3. $$\frac{2}{3-b}$$
4. $$-1$$