Решебник по алгебре 7 класс Мерзляк ФГОС Задание 53

 

\[\boxed{\text{53\ (53).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ 5ax = - 45;\ \ x = 3\]

\[5 \cdot 3x = - 45\]

\[15a = - 45\]

\[a = - \frac{45}{15}\]

\[a = - 3\]

\[Ответ:\ при\ a = - 3.\]

\[2)\ (a - 4) \cdot x = - 5a + 4x - 7;\ \ \ \ \ \]

\[\ x = - 6\]

\[- 6 \cdot (a - 4) = - 5a - 24 - 7\]

\[- 6a + 24 = - 5a - 31\]

\[- 6a + 5a = - 31 - 24\]

\[- a = - 55\]

\[a = 55\]

\[Ответ:при\ a = 55.\]

\[\boxed{\text{53.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ \left( \frac{4^{\backslash 2}}{9} - \frac{5^{\backslash 3}}{6} \right)\ :\left( - \frac{14}{27} \right) =\]

\[= \frac{8 - 15}{18} \cdot \left( - \frac{27}{14} \right) =\]

\[= - \frac{7}{18} \cdot \left( - \frac{27}{14} \right) = \frac{3}{2 \cdot 2} = \frac{3}{4}.\]

\[2) - 1,5 \cdot 4 - 2 = - 6 - 2 = - 8.\]

\[3)\ ( - 1,9 + 0,9)( - 1,9 - 0,9) =\]

\[= - 1 \cdot ( - 2,8) = 2,8.\]

\[4)\ (6 - 8)^{3} = ( - 2)^{3} = - 8.\]