Упр.14.9 ГДЗ Мордкович Семенов 8 класс (Алгебра)

а) (b — 2)/(b + 1) — (1 — 5b)/((b — 1)(b + 1));
б) 3x/(x + 4) — (x^2 — 20x)/((x + 4)(x — 4));
в) (6y + 3)/((2y + 3)(2y — 3)) + 2/(3 — 2y);
г) (18a — 4)/((2 + 3a)(2 — 3a)) + 3a/(3a — 2);
д) 2d/(1 — d^2) + 1/(1 + d);
е) 2p/(p^2 — 9) — 1/(p — 3).

1. а)

   $$\frac{b-2}{b+1}-\frac{1-5b}{(b-1)(b+1)}= \frac{(b-2)(b-1)-(1-5b)}{(b-1)(b+1)}$$
   $$=\frac{b^2-3b+2-1+5b}{(b-1)(b+1)}= \frac{b^2+2b+1}{(b-1)(b+1)}= \frac{(b+1)^2}{(b-1)(b+1)}= \frac{b+1}{b-1}.$$

2. б)

   $$\frac{3x}{x+4}-\frac{x^2-20x}{(x+4)(x-4)}= \frac{3x(x-4)-(x^2-20x)}{(x+4)(x-4)}$$
   $$=\frac{3x^2-12x-x^2+20x}{(x+4)(x-4)}= \frac{2x^2+8x}{(x+4)(x-4)}= \frac{2x(x+4)}{(x+4)(x-4)}= \frac{2x}{x-4}.$$

3. в)

   $$\frac{6y+3}{(2y+3)(2y-3)}+\frac{2}{3-2y}= \frac{6y+3}{(2y+3)(2y-3)}-\frac{2}{2y-3}$$
   $$=\frac{6y+3-2(2y+3)}{(2y+3)(2y-3)}= \frac{6y+3-4y-6}{(2y+3)(2y-3)}= \frac{2y-3}{(2y+3)(2y-3)}= \frac{1}{2y+3}.$$

4. г)

   $$\frac{18a-4}{(2+3a)(2-3a)}+\frac{3a}{3a-2}= \frac{18a-4}{(2+3a)(2-3a)}-\frac{3a}{2-3a}$$
   $$=\frac{18a-4-3a(2+3a)}{(2+3a)(2-3a)}= \frac{18a-4-6a-9a^2}{(2+3a)(2-3a)}$$
   $$=\frac{-9a^2+12a-4}{(2+3a)(2-3a)}= \frac{-(9a^2-12a+4)}{(2+3a)(2-3a)}= \frac{-(3a-2)^2}{(2+3a)(2-3a)}$$
   $$=\frac{-(2-3a)^2}{(2+3a)(2-3a)}= \frac{3a-2}{2+3a}.$$

5. д)

   $$\frac{2d}{1-d^2}+\frac{1}{1+d}= \frac{2d}{(1-d)(1+d)}+\frac{1-d}{(1-d)(1+d)}$$
   $$=\frac{2d+1-d}{(1-d)(1+d)}= \frac{d+1}{(1-d)(1+d)}= \frac{1}{1-d}.$$

6. е)

   $$\frac{2p}{p^2-9}-\frac{1}{p-3}= \frac{2p}{(p-3)(p+3)}-\frac{p+3}{(p-3)(p+3)}$$
   $$=\frac{2p-p-3}{(p-3)(p+3)}= \frac{p-3}{(p-3)(p+3)}= \frac{1}{p+3}.$$

Ответ

а) $$\frac{b+1}{b-1}$$; б) $$\frac{2x}{x-4}$$; в) $$\frac{1}{2y+3}$$; г) $$\frac{3a-2}{2+3a}$$; д) $$\frac{1}{1-d}$$; е) $$\frac{1}{p+3}$$.