Страница 32 номер 1.91, ГДЗ по алгебре за 8 класс к учебнику Дорофеева
Упростите выражение (1.91—1.94).
а) \(\frac{xy}{x-y}\cdot\left(\frac{1}{y^2}-\frac{1}{x^2}\right)\);
б) \(\frac{mn^2}{n^2-m^2}\cdot\left(\frac{2}{m}-\frac{2}{n}\right)\);
в) \(\left(a-\frac{6a-4}{a+2}\right)\cdot\frac{a+2}{a^2-2a}\);
г) \(\left(\frac{u}{u-v}-\frac{u}{u+v}\right)\cdot\frac{u^2+uv}{2v}\);
д) \(\left(\frac{c-d}{d}+\frac{2c}{c-d}\right):\frac{c^2+d^2}{c-d}\);
е) \(\left(\frac{a+b}{a}-\frac{a+b}{b}\right):\frac{a+b}{a^2b^2}\).
а) \[\begin{aligned} &\frac{xy}{x-y}\cdot\left(\frac{1}{y^2}-\frac{1}{x^2}\right) = \frac{xy}{x-y}\cdot\frac{x^2-y^2}{x^2y^2} = {} \\ &= \frac{xy(x-y)(x+y)}{(x-y)x^2y^2} = \frac{x+y}{xy}. \end{aligned}\]
б) \[\begin{aligned} &\frac{mn^2}{n^2-m^2}\cdot\left(\frac{2}{m}-\frac{2}{n}\right) = \frac{mn^2}{n^2-m^2}\cdot\frac{2(n-m)}{mn} = {} \\ &= \frac{mn^2 \cdot 2(n-m)}{(n-m)(n+m)mn} = \frac{2n}{n+m}. \end{aligned}\]
в) \[\begin{aligned} &\left(a-\frac{6a-4}{a+2}\right)\cdot\frac{a+2}{a^2-2a} = \frac{a(a+2)-(6a-4)}{a+2}\cdot\frac{a+2}{a(a-2)} = {} \\ &= \frac{a^2-4a+4}{a+2}\cdot\frac{a+2}{a(a-2)} = \frac{(a-2)^2(a+2)}{(a+2)a(a-2)} = \frac{a-2}{a}. \end{aligned}\]
г) \[\begin{aligned} &\left(\frac{u}{u-v}-\frac{u}{u+v}\right)\cdot\frac{u^2+uv}{2v} = \frac{u(u+v)-u(u-v)}{(u-v)(u+v)}\cdot\frac{u(u+v)}{2v} = {} \\ &= \frac{2uv}{(u-v)(u+v)}\cdot\frac{u(u+v)}{2v} = \frac{u^2}{u-v}. \end{aligned}\]
д) \[\begin{aligned} &\left(\frac{c-d}{d}+\frac{2c}{c-d}\right):\frac{c^2+d^2}{c-d} = \frac{(c-d)^2+2cd}{d(c-d)}\cdot\frac{c-d}{c^2+d^2} = {} \\ &= \frac{c^2+d^2}{d(c-d)}\cdot\frac{c-d}{c^2+d^2} = \frac{1}{d}. \end{aligned}\]
е) \[\begin{aligned} &\left(\frac{a+b}{a}-\frac{a+b}{b}\right):\frac{a+b}{a^2b^2} = \frac{b(a+b)-a(a+b)}{ab}\cdot\frac{a^2b^2}{a+b} = {} \\ &= \frac{(a+b)(b-a)}{ab}\cdot\frac{a^2b^2}{a+b} = ab(b-a) = ab^2-a^2b. \end{aligned}\]
Ответ: а) \(\frac{x+y}{xy}\); б) \(\frac{2n}{n+m}\); в) \(\frac{a-2}{a}\); г) \(\frac{u^2}{u-v}\); д) \(\frac{1}{d}\); е) \(ab^2-a^2b\).
Упростите выражение (1.91—1.94).
а) \(\frac{xy}{x-y}\cdot\left(\frac{1}{y^2}-\frac{1}{x^2}\right)\);
б) \(\frac{mn^2}{n^2-m^2}\cdot\left(\frac{2}{m}-\frac{2}{n}\right)\);
в) \(\left(a-\frac{6a-4}{a+2}\right)\cdot\frac{a+2}{a^2-2a}\);
г) \(\left(\frac{u}{u-v}-\frac{u}{u+v}\right)\cdot\frac{u^2+uv}{2v}\);
д) \(\left(\frac{c-d}{d}+\frac{2c}{c-d}\right):\frac{c^2+d^2}{c-d}\);
е) \(\left(\frac{a+b}{a}-\frac{a+b}{b}\right):\frac{a+b}{a^2b^2}\).
В каждом случае выражение в скобках заменяем одной дробью, а затем выполняем указанное действие с дробями.
а) \[\begin{aligned} &\frac{xy}{x-y}\cdot\left(\frac{1}{y^2}-\frac{1}{x^2}\right) = \frac{xy}{x-y}\cdot\frac{x^2-y^2}{x^2y^2} = {} \\ &= \frac{xy(x-y)(x+y)}{(x-y)x^2y^2} = \frac{x+y}{xy}. \end{aligned}\]
б) \[\begin{aligned} &\frac{mn^2}{n^2-m^2}\cdot\left(\frac{2}{m}-\frac{2}{n}\right) = \frac{mn^2}{n^2-m^2}\cdot\frac{2(n-m)}{mn} = {} \\ &= \frac{mn^2 \cdot 2(n-m)}{(n-m)(n+m)mn} = \frac{2n}{n+m}. \end{aligned}\]
в) \[\begin{aligned} &\left(a-\frac{6a-4}{a+2}\right)\cdot\frac{a+2}{a^2-2a} = \frac{a(a+2)-(6a-4)}{a+2}\cdot\frac{a+2}{a(a-2)} = {} \\ &= \frac{a^2-4a+4}{a+2}\cdot\frac{a+2}{a(a-2)} = \frac{(a-2)^2(a+2)}{(a+2)a(a-2)} = \frac{a-2}{a}. \end{aligned}\]
г) \[\begin{aligned} &\left(\frac{u}{u-v}-\frac{u}{u+v}\right)\cdot\frac{u^2+uv}{2v} = \frac{u(u+v)-u(u-v)}{(u-v)(u+v)}\cdot\frac{u(u+v)}{2v} = {} \\ &= \frac{2uv}{(u-v)(u+v)}\cdot\frac{u(u+v)}{2v} = \frac{u^2}{u-v}. \end{aligned}\]
д) \[\begin{aligned} &\left(\frac{c-d}{d}+\frac{2c}{c-d}\right):\frac{c^2+d^2}{c-d} = \frac{(c-d)^2+2cd}{d(c-d)}\cdot\frac{c-d}{c^2+d^2} = {} \\ &= \frac{c^2+d^2}{d(c-d)}\cdot\frac{c-d}{c^2+d^2} = \frac{1}{d}. \end{aligned}\]
е) \[\begin{aligned} &\left(\frac{a+b}{a}-\frac{a+b}{b}\right):\frac{a+b}{a^2b^2} = \frac{b(a+b)-a(a+b)}{ab}\cdot\frac{a^2b^2}{a+b} = {} \\ &= \frac{(a+b)(b-a)}{ab}\cdot\frac{a^2b^2}{a+b} = ab(b-a) = ab^2-a^2b. \end{aligned}\]
Ответ: а) \(\frac{x+y}{xy}\); б) \(\frac{2n}{n+m}\); в) \(\frac{a-2}{a}\); г) \(\frac{u^2}{u-v}\); д) \(\frac{1}{d}\); е) \(ab^2-a^2b\).
