ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 81

 

\[\boxed{\text{81\ (81).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\mathbf{Знаменатель\ целых\ чисел\ }\]

\[\mathbf{равен\ 1.}\]

\[\textbf{а)}\ 5^{\backslash 2} - \frac{с}{2} = \frac{10 - с}{2}\]

\[\textbf{б)}\ 5{y^{2}}^{\backslash 3} - \frac{15y^{2} - 1}{3} =\]

\[= \frac{15y^{2} - 15y^{2} + 1}{3} = \frac{1}{3}\]

\[\textbf{в)}\ a + b - \frac{a - 3}{3} =\]

\[= (a + b)^{\backslash 3} - \frac{a - 3}{3} =\]

\[= \frac{3a + 3b - a + 3}{3} =\]

\[= \frac{2a + 3b + 3}{3}\]

\[\textbf{г)}\ \frac{2b^{2} - 1}{b} - b + 5 =\]

\[= \ \frac{2b^{2} - 1}{b} - (b - 5)^{\backslash b} =\]

\[= \frac{2b^{2} - 1 - b^{2} + 5b}{b} =\]

\[= \frac{b^{2} + 5b - 1}{b}\]

\[\boxed{\text{81.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \frac{x - y^{\backslash z}}{\text{xy}} - \frac{x - z^{\backslash y}}{\text{xz}} =\]

\[= \frac{zx - zy - yx + yz}{\text{xyz}} =\]

\[= \frac{zx - yx}{\text{xyz}} = \frac{x \cdot (z - y)}{x\text{yz}} = \frac{z - y}{\text{yz}}\]

\[\textbf{б)}\ \frac{a - 2b^{\backslash a}}{3b} - \frac{b - 2a^{\backslash b}}{3a} =\]

\[= \frac{a^{2} - 2ab - b^{2} + 2ab}{3ab} =\]

\[= \frac{a^{2} - b^{2}}{3ab}\]

\[\textbf{в)}\ \frac{p - q^{\backslash q}}{p^{3}q^{2}} - \frac{p + q^{\backslash p}}{p^{2}q^{3}} =\]

\[= \frac{pq - q^{2} - p^{2} - pq}{p^{3}q^{3}} =\]

\[= \frac{- q^{2} - p^{2}}{p^{3}q^{3}} = - \frac{q^{2} + p^{2}}{p^{3}q^{3}}\]

\[\textbf{г)}\ \frac{3m - n^{\backslash 2n}}{3m^{2}n} - \frac{2n - m^{\backslash 3m}}{2mn^{2}} =\]

\[= \frac{6mn - 2n^{2} - 6mn + 3m^{2}}{6{m^{2}n}^{2}} =\]

\[= \frac{- 2n^{2} + 3m^{2}}{6{m^{2}n}^{2}}\]