Страница 32 номер 1.94, ГДЗ по алгебре за 8 класс к учебнику Дорофеева
Упростите выражение (1.91—1.94).
а) \(\left(1-\frac{u}{u+v}\right)\cdot\left(1+\frac{v}{u-v}\right)\);
б) \(\left(\frac{1}{x+z}-\frac{1}{x-z}\right):\left(\frac{1}{x+z}+\frac{1}{x-z}\right)\);
в) \(\left(\frac{a}{b}+\frac{b}{a}+2\right):\left(\frac{a}{b}-\frac{b}{a}\right)\);
г) \(\left(\frac{1}{y}-y\right)\cdot\left(\frac{1}{y+1}-\frac{1}{y-1}\right)\).
а) \[\begin{aligned} &\left(1-\frac{u}{u+v}\right)\cdot\left(1+\frac{v}{u-v}\right) = \frac{u+v-u}{u+v}\cdot\frac{u-v+v}{u-v} = {} \\ &= \frac{v}{u+v}\cdot\frac{u}{u-v} = \frac{uv}{u^2-v^2}. \end{aligned}\]
б) \[\begin{aligned} &\left(\frac{1}{x+z}-\frac{1}{x-z}\right):\left(\frac{1}{x+z}+\frac{1}{x-z}\right) = {} \\ &= \frac{x-z-x-z}{(x+z)(x-z)}:\frac{x-z+x+z}{(x+z)(x-z)} = {} \\ &= \frac{-2z}{(x+z)(x-z)}\cdot\frac{(x+z)(x-z)}{2x} = -\frac{z}{x}. \end{aligned}\]
в) \[\begin{aligned} &\left(\frac{a}{b}+\frac{b}{a}+2\right):\left(\frac{a}{b}-\frac{b}{a}\right) = \frac{a^2+b^2+2ab}{ab}:\frac{a^2-b^2}{ab} = {} \\ &= \frac{(a+b)^2ab}{ab(a-b)(a+b)} = \frac{a+b}{a-b}. \end{aligned}\]
г) \[\begin{aligned} &\left(\frac{1}{y}-y\right)\cdot\left(\frac{1}{y+1}-\frac{1}{y-1}\right) = \frac{1-y^2}{y}\cdot\frac{y-1-y-1}{(y+1)(y-1)} = {} \\ &= \frac{(1-y^2)\cdot(-2)}{y(y^2-1)} = \frac{2(y^2-1)}{y(y^2-1)} = \frac{2}{y}. \end{aligned}\]
Ответ: а) \(\frac{uv}{u^2-v^2}\); б) \(-\frac{z}{x}\); в) \(\frac{a+b}{a-b}\); г) \(\frac{2}{y}\).
Упростите выражение (1.91—1.94).
а) \(\left(1-\frac{u}{u+v}\right)\cdot\left(1+\frac{v}{u-v}\right)\);
б) \(\left(\frac{1}{x+z}-\frac{1}{x-z}\right):\left(\frac{1}{x+z}+\frac{1}{x-z}\right)\);
в) \(\left(\frac{a}{b}+\frac{b}{a}+2\right):\left(\frac{a}{b}-\frac{b}{a}\right)\);
г) \(\left(\frac{1}{y}-y\right)\cdot\left(\frac{1}{y+1}-\frac{1}{y-1}\right)\).
Каждое выражение в скобках заменяем одной дробью, а затем выполняем умножение или деление дробей.
а) \[\begin{aligned} &\left(1-\frac{u}{u+v}\right)\cdot\left(1+\frac{v}{u-v}\right) = \frac{u+v-u}{u+v}\cdot\frac{u-v+v}{u-v} = {} \\ &= \frac{v}{u+v}\cdot\frac{u}{u-v} = \frac{uv}{u^2-v^2}. \end{aligned}\]
б) \[\begin{aligned} &\left(\frac{1}{x+z}-\frac{1}{x-z}\right):\left(\frac{1}{x+z}+\frac{1}{x-z}\right) = {} \\ &= \frac{x-z-x-z}{(x+z)(x-z)}:\frac{x-z+x+z}{(x+z)(x-z)} = {} \\ &= \frac{-2z}{(x+z)(x-z)}\cdot\frac{(x+z)(x-z)}{2x} = -\frac{z}{x}. \end{aligned}\]
в) \[\begin{aligned} &\left(\frac{a}{b}+\frac{b}{a}+2\right):\left(\frac{a}{b}-\frac{b}{a}\right) = \frac{a^2+b^2+2ab}{ab}:\frac{a^2-b^2}{ab} = {} \\ &= \frac{(a+b)^2ab}{ab(a-b)(a+b)} = \frac{a+b}{a-b}. \end{aligned}\]
г) \[\begin{aligned} &\left(\frac{1}{y}-y\right)\cdot\left(\frac{1}{y+1}-\frac{1}{y-1}\right) = \frac{1-y^2}{y}\cdot\frac{y-1-y-1}{(y+1)(y-1)} = {} \\ &= \frac{(1-y^2)\cdot(-2)}{y(y^2-1)} = \frac{2(y^2-1)}{y(y^2-1)} = \frac{2}{y}. \end{aligned}\]
Ответ: а) \(\frac{uv}{u^2-v^2}\); б) \(-\frac{z}{x}\); в) \(\frac{a+b}{a-b}\); г) \(\frac{2}{y}\).
