Задание 8 Параграф 4 ГДЗ Рабочая тетрадь 1 Мерзляк Полонский 8 класс (Алгебра)

Задача

1) (a^2+4)/(a^2-4a+4)+(a+2)/(a-2);
2) (6c+15)/(36c^2-1)-(3c+4)/(18c^2+3c);
3) 2/3x-(6x+32)/(64-9x^2 )-4/(3x-8);
4) m/(m+3)-1-(15-m)/(m^2-9);
5) a^2-6a+36-(a^3-216)/(a+6);
6) 2/(2b-5)^2 -3/(2b+5)^2 +1/(4b^2-25).

Подробный ответ

$$\frac{a^2+4}{a^2-4a+4}+\frac{a+2}{a-2}=\frac{a^2+4}{(a-2)^2}+\frac{a+2}{a-2}$$
$$=\frac{a^2+4+(a+2)(a-2)}{(a-2)^2}=\frac{a^2+4+a^2-4}{(a-2)^2}=\frac{2a^2}{(a-2)^2}$$
$$\frac{6c+15}{36c^2-1}-\frac{3c+4}{18c^2+3c}=\frac{6c+15}{(6c-1)(6c+1)}-\frac{3c+4}{3c(6c+1)}$$
$$=\frac{3c(6c+15)-(3c+4)(6c-1)}{3c(6c-1)(6c+1)}$$
$$=\frac{18c^2+45c-18c^2+3c-24c+4}{3c(6c-1)(6c+1)}=\frac{24c+4}{3c(6c-1)(6c+1)}$$
$$=\frac{4(6c+1)}{3c(6c-1)(6c+1)}=\frac{4}{3c(6c-1)}$$
$$\frac{2}{3x}-\frac{6x+32}{64-9x^2}-\frac{4}{3x-8}=\frac{2}{3x}-\frac{6x+32}{(8-3x)(8+3x)}+\frac{4}{8-3x}$$
$$=\frac{2(8-3x)(8+3x)-3x(6x+32)+4\cdot 3x(8+3x)}{3x(8-3x)(8+3x)}$$
$$=\frac{2(64-9x^2)-18x^2-96x+96x+36x^2}{3x(64-9x^2)}=\frac{128}{3x(64-9x^2)}$$
$$\frac{m}{m+3}-1-\frac{15-m}{m^2-9}=\frac{m}{m+3}-1-\frac{15-m}{(m-3)(m+3)}$$
$$=\frac{m(m-3)-(m^2-9)-(15-m)}{(m-3)(m+3)}$$
$$=\frac{m^2-3m-m^2+9-15+m}{m^2-9}=\frac{-2m-6}{m^2-9}$$
$$=\frac{-2(m+3)}{(m-3)(m+3)}=-\frac{2}{m-3}=\frac{2}{3-m}$$
$$a^2-6a+36-\frac{a^3-216}{a+6}=\frac{(a^2-6a+36)(a+6)-(a^3-216)}{a+6}$$
$$=\frac{a^3+216-a^3+216}{a+6}=\frac{432}{a+6}$$
$$\frac{2}{(2b-5)^2}-\frac{3}{(2b+5)^2}+\frac{1}{4b^2-25}=\frac{2}{(2b-5)^2}-\frac{3}{(2b+5)^2}+\frac{1}{(2b-5)(2b+5)}$$
$$=\frac{2(2b+5)^2-3(2b-5)^2+(2b-5)(2b+5)}{(2b-5)^2(2b+5)^2}$$
$$=\frac{2(4b^2+20b+25)-3(4b^2-20b+25)+4b^2-25}{(2b-5)^2(2b+5)^2}$$
$$=\frac{8b^2+40b+50-12b^2+60b-75+4b^2-25}{(2b-5)^2(2b+5)^2}$$
$$=\frac{100b-50}{(2b-5)^2(2b+5)^2}=\frac{50(2b-1)}{(2b-5)^2(2b+5)^2}$$