Решебник по алгебре 8 класс Мерзляк Задание 110

\[\boxed{\text{110\ (110).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\frac{a}{b} + 1^{\backslash b} = \frac{a + b}{b}\]

\[2)\frac{x}{y} - x^{\backslash y} = \frac{x - xy}{y}\]

\[3)\frac{m^{\backslash m}}{n} + \frac{n^{\backslash n}}{m} + 2^{\backslash mn} =\]

\[= \frac{m^{2} + n^{2} + 2nm}{\text{nm}} = \frac{(m + n)^{2}}{\text{mn}}\]

\[4)\frac{9}{p^{2}} - \frac{4^{\backslash p}}{p} + 3^{\backslash p^{2}} = \frac{9 - 4p + 3p^{2}}{p^{2}}\]

\[5)\ 2^{\backslash a} - \frac{3b + 2a}{a} =\]

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

\[6)\frac{3b + 4}{b - 2} - 3^{\backslash b - 2} =\]

\[= \frac{3b + 4 - 3 \cdot (b - 2)}{b - 2} =\]

\[= \frac{3b + 4 - 3b + 6}{b - 2} = \frac{10}{b - 2}\]

\[7)\ 6m^{\backslash 2m} - \frac{12m^{2} + 1}{2m} =\]

\[= \frac{6m \cdot 2m - 12m^{2} - 1}{2m} =\]

\[= \frac{12m^{2} - 12m^{2} - 1}{2m} = - \frac{1}{2m}\]

\[8)\frac{20b^{2} + 5}{2b - 1} - 10b^{\backslash 2b - 1} =\]

\[= \frac{20b^{2} + 5 - 10b \cdot (2b - 1)}{2b - 1} =\]

\[= \frac{20b^{2} + 5 - 20b^{2} + 10b}{2b - 1} =\]

\[= \frac{10b + 5}{2b - 1}\ \]