Решебник по алгебре 8 класс Мерзляк ФГОС Задание 96

 

\[\boxed{\text{96\ (96).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ \frac{1}{3a} = \frac{b}{3ab}\]

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

\[2)\ \frac{4m}{p^{3}q^{2}} = \frac{4mq}{p^{3}q^{3}}\]

\[\frac{3n}{p^{2}q^{3}} = \frac{3np}{p^{3}q^{3}}\ \]

\[3)\ \frac{5}{m - n} = \frac{5 \cdot (m + n)}{(m - n)(m + n)} =\]

\[= \frac{5m + 5n}{(m - n)(m + n)}\]

\[\frac{6}{m + n} = \frac{6 \cdot (m - n)}{(m - n)(m + n)} =\]

\[= \frac{6m - 6n}{(m - n)(m + n)}\]

\[4)\ \frac{6x}{x - 2y} = \frac{6x(x + y)}{(x - 2y)(x + y)}\]

\[\frac{y}{x + y} = \frac{y(x - 2y)}{(x - 2y)(x + y)}\]

\[5)\ \frac{y}{6y - 36} = \frac{y \cdot y}{6y(y - 6)} =\]

\[= \frac{y^{2}}{6y(y - 6)}\]

\[\frac{1}{y^{2} - 6y} = \frac{1 \cdot 6}{6 \cdot y \cdot (y - 6)} =\]

\[= \frac{6}{6y(y - 6)}\]

\[6)\ \frac{1}{a^{2} - 1} = \frac{1 \cdot a}{(a - 1)(a + 1) \cdot a} =\]

\[= \frac{a}{a(a^{2} - 1)}\]

\[\frac{1}{a^{2} + a} = \frac{1 \cdot (a - 1)}{a \cdot (a + 1)(a - 1)} =\]

\[= \frac{a - 1}{a(a^{2} - 1)}\]

\[\boxed{\text{96.\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[1)\ \frac{x^{\backslash 3}}{4} + \frac{2x^{\backslash 4}}{3} = \frac{3x + 8x}{12} = \frac{11x}{12}\]

\[2)\ \frac{5b}{14} - \frac{b^{\backslash 2}}{7} = \frac{5b - 2b}{14} = \frac{3b}{14}\]

\[3)\ \frac{m^{\backslash 3}}{8} - \frac{n^{\backslash 4}}{6} = \frac{3m - 4n}{24}\]

\[4)\ \frac{4^{\backslash y}}{x} - \frac{3^{\backslash x}}{y} = \frac{4y - 3x}{\text{xy}}\]

\[5)\ \frac{m^{\backslash 3}}{4n} + \frac{m^{\backslash 2}}{6n} = \frac{3m + 2m}{12n} = \frac{5m}{12n}\]

\[6)\ \frac{c^{\backslash 3}}{b} - \frac{d}{3b} = \frac{3c - d}{3b}\]

\[7)\ \frac{a^{\backslash ab^{2}}}{b^{2}} + \frac{1}{ab^{4}} = \frac{a^{2}b^{2} + 1}{ab^{4}}\]

\[8)\ \frac{11^{\backslash 3b}}{5a} - \frac{2c}{15ab} = \frac{33b - 2c}{15ab}\]

\[9)\ \frac{m^{\backslash m}}{\text{abc}} + \frac{c^{\backslash c}}{\text{abm}} = \frac{m^{2} + c^{2}}{\text{abcm}}\]