Задание 6 Параграф 6 ГДЗ Рабочая тетрадь 1 Мерзляк Полонский 8 класс (Алгебра)

1) (c+6)/(c^2-10c+25):(c^2-36)/(4c-20)-4/(c-6);
2) ((m+3)/(m-3)+(m-3)/(m+3)):(4m^2+36)/(m^2+6m+9);
3) (a+b-2ab/(a+b)):(a^2+b^2)/(a^2-b^2 );
4) (3c/(c-2)-c):(c-(8c-25)/(c-2));
5) (7/(a^2-7a)-2/(a-7)-a/(49-7a)):(a^2-49)/a;
6) ((a-2)/(a^2-2a+4)-(6a-13)/(a^3+8)):(a-3)/(5a^3+40).

1. $$\frac{c+6}{c^2-10c+25}:\frac{c^2-36}{4c-20}-\frac{4}{c-6}$$
   $$=\frac{c+6}{(c-5)^2}:\frac{(c-6)(c+6)}{4(c-5)}-\frac{4}{c-6}$$
   $$=\frac{c+6}{(c-5)^2}\cdot\frac{4(c-5)}{(c-6)(c+6)}-\frac{4}{c-6}$$
   $$=\frac{4}{(c-5)(c-6)}-\frac{4}{c-6}$$
   $$=\frac{4-4(c-5)}{(c-5)(c-6)}=\frac{24-4c}{(c-5)(c-6)}$$
   $$=\frac{-4(c-6)}{(c-5)(c-6)}=-\frac{4}{c-5}=\frac{4}{5-c}.$$
2. $$\left(\frac{m+3}{m-3}+\frac{m-3}{m+3}\right):\frac{4m^2+36}{m^2+6m+9}$$
   $$=\frac{(m+3)^2+(m-3)^2}{(m-3)(m+3)}:\frac{4(m^2+9)}{(m+3)^2}$$
   $$=\frac{2m^2+18}{(m-3)(m+3)}\cdot\frac{(m+3)^2}{4(m^2+9)}$$
   $$=\frac{2(m^2+9)}{(m-3)(m+3)}\cdot\frac{(m+3)^2}{4(m^2+9)}$$
   $$=\frac{m+3}{2(m-3)}.$$
3. $$\left(a+b-\frac{2ab}{a+b}\right):\frac{a^2+b^2}{a^2-b^2}$$
   $$=\frac{(a+b)^2-2ab}{a+b}:\frac{a^2+b^2}{(a-b)(a+b)}$$
   $$=\frac{a^2+b^2}{a+b}\cdot\frac{(a-b)(a+b)}{a^2+b^2}=a-b.$$
4. $$\left(\frac{3c}{c-2}-c\right):\left(c-\frac{8c-25}{c-2}\right)$$
   $$=\frac{3c-c(c-2)}{c-2}:\frac{c(c-2)-(8c-25)}{c-2}$$
   $$=\frac{3c-c^2+2c}{c-2}:\frac{c^2-10c+25}{c-2}$$
   $$=\frac{c(5-c)}{c-2}:\frac{(c-5)^2}{c-2}$$
   $$=\frac{c(5-c)}{c-2}\cdot\frac{c-2}{(c-5)^2}=\frac{c}{5-c}.$$
5. $$\left(\frac{7}{a^2-7a}-\frac{2}{a-7}-\frac{a}{49-7a}\right):\frac{a^2-49}{a}$$
   $$=\left(\frac{7}{a(a-7)}-\frac{2}{a-7}+\frac{a}{7(a-7)}\right):\frac{a^2-49}{a}$$
   $$=\frac{49-14a+a^2}{7a(a-7)}:\frac{a^2-49}{a}$$
   $$=\frac{(a-7)^2}{7a(a-7)}:\frac{(a-7)(a+7)}{a}$$
   $$=\frac{a-7}{7a}\cdot\frac{a}{(a-7)(a+7)}=\frac{1}{7(a+7)}.$$
6. $$\left(\frac{a-2}{a^2-2a+4}-\frac{6a-13}{a^3+8}\right):\frac{a-3}{5a^3+40}$$
   $$=\left(\frac{a-2}{a^2-2a+4}-\frac{6a-13}{(a+2)(a^2-2a+4)}\right):\frac{a-3}{5(a^3+8)}$$
   $$=\frac{(a-2)(a+2)-(6a-13)}{(a+2)(a^2-2a+4)}:\frac{a-3}{5(a^3+8)}$$
   $$=\frac{a^2-4-6a+13}{a^3+8}:\frac{a-3}{5(a^3+8)}$$
   $$=\frac{a^2-6a+9}{a^3+8}\cdot\frac{5(a^3+8)}{a-3}$$
   $$=\frac{(a-3)^2}{a^3+8}\cdot\frac{5(a^3+8)}{a-3}=5(a-3).$$

Ответ

1) $$\frac{4}{5-c}$$; 2) $$\frac{m+3}{2(m-3)}$$; 3) $$a-b$$; 4) $$\frac{c}{5-c}$$; 5) $$\frac{1}{7(a+7)}$$; 6) $$5(a-3)$$.