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

$$\frac{c+6}{c^2-10c+25}:\frac{c^2-36}{4c-20}-\frac{4}{c-6}$$
$$=\frac{c+6}{(c-5)^2}:\frac{(c-6)(c+6)}{4(c-5)}-\frac{4}{c-6}$$
$$=\frac{c+6}{(c-5)^2}\cdot\frac{4(c-5)}{(c-6)(c+6)}-\frac{4}{c-6}$$
$$=\frac{4}{(c-5)(c-6)}-\frac{4}{c-6}$$
$$=\frac{4-4(c-5)}{(c-5)(c-6)}=\frac{24-4c}{(c-5)(c-6)}$$
$$=\frac{-4(c-6)}{(c-5)(c-6)}=-\frac{4}{c-5}=\frac{4}{5-c}.$$

$$\left(\frac{m+3}{m-3}+\frac{m-3}{m+3}\right):\frac{4m^2+36}{m^2+6m+9}$$
$$=\frac{(m+3)^2+(m-3)^2}{(m-3)(m+3)}:\frac{4(m^2+9)}{(m+3)^2}$$
$$=\frac{2m^2+18}{(m-3)(m+3)}\cdot\frac{(m+3)^2}{4(m^2+9)}$$
$$=\frac{2(m^2+9)}{(m-3)(m+3)}\cdot\frac{(m+3)^2}{4(m^2+9)}$$
$$=\frac{m+3}{2(m-3)}.$$

$$\left(a+b-\frac{2ab}{a+b}\right):\frac{a^2+b^2}{a^2-b^2}$$
$$=\frac{(a+b)^2-2ab}{a+b}:\frac{a^2+b^2}{(a-b)(a+b)}$$
$$=\frac{a^2+b^2}{a+b}\cdot\frac{(a-b)(a+b)}{a^2+b^2}=a-b.$$

$$\left(\frac{3c}{c-2}-c\right):\left(c-\frac{8c-25}{c-2}\right)$$
$$=\frac{3c-c(c-2)}{c-2}:\frac{c(c-2)-(8c-25)}{c-2}$$
$$=\frac{3c-c^2+2c}{c-2}:\frac{c^2-10c+25}{c-2}$$
$$=\frac{5c-c^2}{c-2}:\frac{(c-5)^2}{c-2}$$
$$=\frac{c(5-c)}{c-2}\cdot\frac{c-2}{(c-5)^2}=\frac{c}{5-c}.$$

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

$$\left(\frac{a-2}{a^2-2a+4}-\frac{6a-13}{a^3+8}\right):\frac{a-3}{5a^3+40}$$
$$=\left(\frac{a-2}{a^2-2a+4}-\frac{6a-13}{(a+2)(a^2-2a+4)}\right):\frac{a-3}{5(a^3+8)}$$
$$=\frac{(a-2)(a+2)-(6a-13)}{(a+2)(a^2-2a+4)}:\frac{a-3}{5(a^3+8)}$$
$$=\frac{a^2-4-6a+13}{a^3+8}:\frac{a-3}{5(a^3+8)}$$
$$=\frac{a^2-6a+9}{a^3+8}\cdot\frac{5(a^3+8)}{a-3}$$
$$=\frac{(a-3)^2\cdot 5(a^3+8)}{(a^3+8)(a-3)}=5(a-3).$$