Упр.25 Вариант 2 Дидактические материалы ГДЗ Мерзляк Полонский 8 класс (Алгебра)
1) (x+4)/(2x-6)-(x+1)/(x-3); 4) 4b/(3b-21)+3b/(14-2b);
2) (a+3)/(3a-3)+(2-a)/(5a-5); 5) 3p/(3p+2q)-(9p^2)/(9p^2+12pq+4q^2);
3) (x+5)/(x-5)-(x-1)/(x+5); 6) 4/(c^2-36)-2/(c^2-6c).
$$\frac{x+4}{2x-6}-\frac{x+1}{x-3}=\frac{x+4}{2(x-3)}-\frac{x+1}{x-3}$$
$$=\frac{x+4-2(x+1)}{2(x-3)}=\frac{x+4-2x-2}{2(x-3)}=\frac{2-x}{2(x-3)}$$
$$\frac{a+3}{3a-3}+\frac{2-a}{5a-5}=\frac{a+3}{3(a-1)}+\frac{2-a}{5(a-1)}$$
$$=\frac{5(a+3)+3(2-a)}{15(a-1)}=\frac{5a+15+6-3a}{15(a-1)}=\frac{2a+21}{15(a-1)}$$
$$\frac{x+5}{x-5}-\frac{x-1}{x+5}=\frac{(x+5)^2-(x-1)(x-5)}{(x-5)(x+5)}$$
$$=\frac{x^2+10x+25-(x^2-6x+5)}{x^2-25}=\frac{16x+20}{x^2-25}$$
$$\frac{4b}{3b-21}+\frac{3b}{14-2b}=\frac{4b}{3(b-7)}-\frac{3b}{2(b-7)}$$
$$=\frac{8b-9b}{6(b-7)}=\frac{-b}{6(b-7)}=\frac{b}{6(7-b)}$$
$$\frac{3p}{3p+2q}-\frac{9p^2}{9p^2+12pq+4q^2}=\frac{3p}{3p+2q}-\frac{9p^2}{(3p+2q)^2}$$
$$=\frac{3p(3p+2q)-9p^2}{(3p+2q)^2}=\frac{9p^2+6pq-9p^2}{(3p+2q)^2}=\frac{6pq}{(3p+2q)^2}$$
$$\frac{4}{c^2-36}-\frac{2}{c^2-6c}=\frac{4}{(c-6)(c+6)}-\frac{2}{c(c-6)}$$
$$=\frac{4c-2(c+6)}{c(c-6)(c+6)}=\frac{4c-2c-12}{c(c-6)(c+6)}=\frac{2(c-6)}{c(c-6)(c+6)}$$
$$=\frac{2}{c(c+6)}$$
Ответ
1) $$\frac{2-x}{2(x-3)}$$; 2) $$\frac{2a+21}{15(a-1)}$$; 3) $$\frac{16x+20}{x^2-25}$$; 4) $$\frac{b}{6(7-b)}$$; 5) $$\frac{6pq}{(3p+2q)^2}$$; 6) $$\frac{2}{c(c+6)}$$.
