Решебник по алгебре 8 класс Мерзляк Задание 98

\[\boxed{\text{98\ (98).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

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

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

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

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

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

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

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

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

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