Упр.16.2 ГДЗ Мордкович Семенов 8 класс (Алгебра)

Задача

а) (p — 5)/(p + 6) : (p — 11p/(p + 6));
б) (q/(q — 5) — 2q) : (11 — 2q)/(q — 5);
в) (cd — d^2)/(c^2 + d^2) · (c/(c + d) + d/(c — d));
г) (6/(x — y) — 5/(x + y)) · (x — y)/(x + 11y);
д) (x^2 — 9)/(2x^2 + 1) · ((6x + 1)/(x — 3) + (6x — 1)/(x + 3));
е) (b + 3)/(b^3 + 9b) · ((b + 3)/(b — 3) + (b — 3)/(b + 3)).

Подробный ответ

а)
$$\frac{p-5}{p+6}:\left(p-\frac{11p}{p+6}\right)=\frac{p-5}{p+6}:\frac{p(p+6)-11p}{p+6}$$
$$=\frac{p-5}{p+6}:\frac{p^2-5p}{p+6}=\frac{p-5}{p+6}\cdot\frac{p+6}{p(p-5)}=\frac{1}{p}.$$
б)
$$\left(\frac{q}{q-5}-2q\right):\frac{11-2q}{q-5}=\frac{q-2q(q-5)}{q-5}:\frac{11-2q}{q-5}$$
$$=\frac{q-2q^2+10q}{q-5}:\frac{11-2q}{q-5}=\frac{11q-2q^2}{q-5}:\frac{11-2q}{q-5}$$
$$=\frac{q(11-2q)}{q-5}\cdot\frac{q-5}{11-2q}=q.$$
в)
$$\frac{cd-d^2}{c^2+d^2}\cdot\left(\frac{c}{c+d}+\frac{d}{c-d}\right)=\frac{cd-d^2}{c^2+d^2}\cdot\frac{c(c-d)+d(c+d)}{(c-d)(c+d)}$$
$$=\frac{cd-d^2}{c^2+d^2}\cdot\frac{c^2-dc+dc+d^2}{(c-d)(c+d)}=\frac{cd-d^2}{c^2+d^2}\cdot\frac{c^2+d^2}{(c-d)(c+d)}$$
$$=\frac{d(c-d)}{(c-d)(c+d)}=\frac{d}{c+d}.$$
г)
$$\left(\frac{6}{x-y}-\frac{5}{x+y}\right)\cdot\frac{x-y}{x+11y}=\frac{6(x+y)-5(x-y)}{(x-y)(x+y)}\cdot\frac{x-y}{x+11y}$$
$$=\frac{6x+6y-5x+5y}{(x-y)(x+y)}\cdot\frac{x-y}{x+11y}=\frac{x+11y}{(x-y)(x+y)}\cdot\frac{x-y}{x+11y}=\frac{1}{x+y}.$$
д)
$$\frac{x^2-9}{2x^2+1}\cdot\left(\frac{6x+1}{x-3}+\frac{6x-1}{x+3}\right)=\frac{x^2-9}{2x^2+1}\cdot\frac{(6x+1)(x+3)+(6x-1)(x-3)}{(x-3)(x+3)}$$
$$=\frac{x^2-9}{2x^2+1}\cdot\frac{6x^2+18x+x+3+6x^2-18x-x+3}{(x-3)(x+3)}$$
$$=\frac{x^2-9}{2x^2+1}\cdot\frac{12x^2+6}{(x-3)(x+3)}=\frac{x^2-9}{2x^2+1}\cdot\frac{6(2x^2+1)}{x^2-9}=6.$$
е)
$$\frac{b+3}{b^3+9b}\cdot\left(\frac{b+3}{b-3}+\frac{b-3}{b+3}\right)=\frac{b+3}{b(b^2+9)}\cdot\frac{(b+3)^2+(b-3)^2}{(b-3)(b+3)}$$
$$=\frac{b+3}{b(b^2+9)}\cdot\frac{b^2+6b+9+b^2-6b+9}{(b-3)(b+3)}$$
$$=\frac{b+3}{b(b^2+9)}\cdot\frac{2b^2+18}{(b-3)(b+3)}=\frac{b+3}{b(b^2+9)}\cdot\frac{2(b^2+9)}{(b-3)(b+3)}$$
$$=\frac{2}{b(b-3)}=\frac{2}{b^2-3b}.$$