Упр.14.11 ГДЗ Мордкович Семенов 8 класс (Алгебра)

Задача

а) (x — 2y)/(xy + y^2) + (2x — y)/(x^2 + xy);
б) (3a + b)/(a^2 — ab) — (a + 3b)/(ab — b^2);
в) (1 — x)/(x^2 — xy) — (y — 1)/(y^2 — xy);
г) (3 + c)/(c^2 — cd) + (3 + d)/(d^2 — cd);
д) (p — q)/(2p^2 + 2pq) + 2q/(p^2 — q^2);
е) (3m + n)/(9m^2 — 3mn) — 4n/(9m^2 — n^2).

Подробный ответ

а)
$$\frac{x-2y}{xy+y^2}+\frac{2x-y}{x^2+xy} =\frac{x-2y}{y(x+y)}+\frac{2x-y}{x(x+y)}$$
$$=\frac{x(x-2y)+y(2x-y)}{xy(x+y)} =\frac{x^2-2xy+2xy-y^2}{xy(x+y)}$$
$$=\frac{x^2-y^2}{xy(x+y)} =\frac{(x-y)(x+y)}{xy(x+y)} =\frac{x-y}{xy}.$$
б)
$$\frac{3a+b}{a^2-ab}-\frac{a+3b}{ab-b^2} =\frac{3a+b}{a(a-b)}-\frac{a+3b}{b(a-b)}$$
$$=\frac{b(3a+b)-a(a+3b)}{ab(a-b)} =\frac{3ab+b^2-a^2-3ab}{ab(a-b)}$$
$$=\frac{b^2-a^2}{ab(a-b)} =\frac{(b-a)(b+a)}{ab(a-b)} =-\frac{a+b}{ab}.$$
в)
$$\frac{1-x}{x^2-xy}-\frac{y-1}{y^2-xy} =\frac{1-x}{x(x-y)}-\frac{y-1}{y(y-x)}$$
$$=\frac{1-x}{x(x-y)}+\frac{y-1}{y(x-y)} =\frac{y(1-x)+x(y-1)}{xy(x-y)}$$
$$=\frac{y-xy+xy-x}{xy(x-y)} =\frac{y-x}{xy(x-y)} =-\frac{1}{xy}.$$
г)
$$\frac{3+c}{c^2-cd}+\frac{3+d}{d^2-cd} =\frac{3+c}{c(c-d)}+\frac{3+d}{d(d-c)}$$
$$=\frac{d(3+c)-c(3+d)}{cd(c-d)} =\frac{3d+cd-3c-cd}{cd(c-d)}$$
$$=\frac{3d-3c}{cd(c-d)} =\frac{-3(c-d)}{cd(c-d)} =-\frac{3}{cd}.$$
д)
$$\frac{p-q}{2p^2+2pq}+\frac{2q}{p^2-q^2} =\frac{p-q}{2p(p+q)}+\frac{2q}{(p-q)(p+q)}$$
$$=\frac{(p-q)^2+4pq}{2p(p-q)(p+q)} =\frac{p^2-2pq+q^2+4pq}{2p(p-q)(p+q)}$$
$$=\frac{p^2+2pq+q^2}{2p(p-q)(p+q)} =\frac{(p+q)^2}{2p(p-q)(p+q)} =\frac{p+q}{2p(p-q)}.$$
е)
$$\frac{3m+n}{9m^2-3mn}-\frac{4n}{9m^2-n^2} =\frac{3m+n}{3m(3m-n)}-\frac{4n}{(3m-n)(3m+n)}$$
$$=\frac{(3m+n)^2-12mn}{3m(3m-n)(3m+n)}$$
$$=\frac{9m^2+6mn+n^2-12mn}{3m(3m-n)(3m+n)}$$
$$=\frac{9m^2-6mn+n^2}{3m(3m-n)(3m+n)} =\frac{(3m-n)^2}{3m(3m-n)(3m+n)}$$
$$=\frac{3m-n}{3m(3m+n)}.$$