Упр.14.8 ГДЗ Мордкович Семенов 8 класс (Алгебра)
а) (x — 2)/(x + 2) — (x + 3)/(x — 3);
б) (p + 2)/(p + 4) — (p + 3)/(p + 6);
в) (b — c)/(4bc(b + c)) — (b + c)/(4bc(b — c));
г) (d — c)/(8c^2 (3c — d)) + (3c + d)/(8c^2 (d + c));
д) (5x — 2)/((x — 1)(x + 2)) + (x — 2)/(x + 2);
е) (a — 3)/(a + 3) + (8a — 6)/((a — 2)(a + 3)).
а)
$$\frac{x-2}{x+2}-\frac{x+3}{x-3}= \frac{(x-2)(x-3)-(x+3)(x+2)}{(x+2)(x-3)}$$
$$=\frac{x^2-5x+6-(x^2+5x+6)}{(x+2)(x-3)}= \frac{-10x}{(x+2)(x-3)}$$
б)
$$\frac{p+2}{p+4}-\frac{p+3}{p+6}= \frac{(p+2)(p+6)-(p+3)(p+4)}{(p+4)(p+6)}$$
$$=\frac{p^2+8p+12-(p^2+7p+12)}{(p+4)(p+6)}= \frac{p}{(p+4)(p+6)}$$
в)
$$\frac{b-c}{4bc(b+c)}-\frac{b+c}{4bc(b-c)}= \frac{(b-c)^2-(b+c)^2}{4bc(b-c)(b+c)}$$
$$=\frac{b^2-2bc+c^2-b^2-2bc-c^2}{4bc(b^2-c^2)}= \frac{-4bc}{4bc(b^2-c^2)}= -\frac{1}{b^2-c^2}$$
$$=\frac{1}{c^2-b^2}$$
г)
$$\frac{d-c}{8c^2(3c-d)}+\frac{3c+d}{8c^2(d+c)}= \frac{(d-c)(d+c)+(3c+d)(3c-d)}{8c^2(3c-d)(d+c)}$$
$$=\frac{d^2-c^2+9c^2-d^2}{8c^2(3c-d)(d+c)}= \frac{8c^2}{8c^2(3c-d)(d+c)}= \frac{1}{(3c-d)(d+c)}$$
д)
$$\frac{5x-2}{(x-1)(x+2)}+\frac{x-2}{x+2}= \frac{5x-2+(x-2)(x-1)}{(x-1)(x+2)}$$
$$=\frac{5x-2+x^2-3x+2}{(x-1)(x+2)}= \frac{x^2+2x}{(x-1)(x+2)}= \frac{x(x+2)}{(x-1)(x+2)}= \frac{x}{x-1}$$
е)
$$\frac{a-3}{a+3}+\frac{8a-6}{(a-2)(a+3)}= \frac{(a-3)(a-2)+8a-6}{(a-2)(a+3)}$$
$$=\frac{a^2-5a+6+8a-6}{(a-2)(a+3)}= \frac{a^2+3a}{(a-2)(a+3)}= \frac{a(a+3)}{(a-2)(a+3)}= \frac{a}{a-2}$$
Ответ
а) $$\frac{-10x}{(x+2)(x-3)}$$; б) $$\frac{p}{(p+4)(p+6)}$$; в) $$\frac{1}{c^2-b^2}$$; г) $$\frac{1}{(3c-d)(d+c)}$$; д) $$\frac{x}{x-1}$$; е) $$\frac{a}{a-2}$$.
