Решебник по алгебре 7 класс Мерзляк ФГОС Задание 126

 

\[\boxed{\text{126\ (126).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[14 - 6 \cdot 4 = 14 - 24 = - 10.\]

\[14 - 6 \cdot ( - 2) = 14 + 12 = 26.\]

\[14 - 6 \cdot 0 = 14 - 0 = 14.\]

\[14 - 6 \cdot ( - 0,3) = 14 + 1,8 =\]

\[= 15,8.\]

\[14 - 6 \cdot \frac{3}{8} = 14 - \frac{6 \cdot 3}{8} =\]

\[= 14 - \frac{3 \cdot 3}{4} = 14 - \frac{9}{4} =\]

\[= 14 - 2\frac{1}{4} = 11\frac{3}{4}.\]

\[7^{2} + 3 = 49 + 3 = 52.\]

\[( - 2)^{2} + 3 = 4 + 3 = 7.\]

\[0^{2} + 3 = 0 + 3 = 3.\]

\[(0,4)^{2} + 3 = 0,16 + 3 = 3,16.\]

\[\left( 1\frac{1}{3} \right)^{2} + 3 = \left( \frac{4}{3} \right)^{2} + 3 =\]

\[= \frac{16}{9} + 3 = 1\frac{7}{9} + 3 = 4\frac{7}{9}.\]

\[(2 \cdot 0,2 - 1) \cdot ( - 0,6) =\]

\[= (0,4 - 1) \cdot ( - 0,6) =\]

\[= - 0,6 \cdot ( - 0,6) = 0,36.\ \ \]

\[\boxed{\text{126.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

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

\[15a = - 45\]

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

\[a = - 3\]

\[Ответ:\ при\ a = - 3.\]

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

\[\ x = - 6\]

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

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

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

\[- a = - 55\]

\[a = 55\]

\[Ответ:при\ a = 55.\]