Решебник по алгебре 7 класс Мерзляк ФГОС Задание 9

 

\[\boxed{\text{9\ (9).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ 0,4\ y + 1\]

\[при\ \ y = - 0,5;8;\ - 10:\]

\[0,4 \cdot ( - 0,5) + 1 = - 0,2 + 1 =\]

\[= 0,8\]

\[0,4 \cdot 8 + 1 = 3,2 + 1 = 4,2\]

\[0,4 \cdot ( - 10) + 1 = - 4 + 1 = - 3.\]

\[2)\ \frac{2}{7}c - 0,2d\]

\[при\ c = - 28,\ d = 15:\]

\[\frac{2}{7} \cdot ( - 28) - 0,2 \cdot 15 =\]

\[= - 2 \cdot 4 - 3 = - 8 - 3 = - 11.\]

\[\boxed{\text{9.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ \frac{4^{\backslash 4}}{15} + \frac{3^{\backslash 15}}{4} = \frac{16 + 45}{60} =\]

\[= \frac{61}{60} = 1\frac{1}{60}\]

\[2)\ \frac{7^{\backslash 3}}{12} - \frac{5^{\backslash 4}}{9} = \frac{21 - 20}{36} = \frac{1}{36}\]

\[3)\ \frac{5}{8} \cdot \frac{4}{15} = \frac{1}{2 \cdot 3} = \frac{1}{6}\]

\[4)\ \frac{13}{17} \cdot 34 = 13 \cdot 2 = 26\]

\[5)\ \frac{12}{35}\ :\frac{3}{5} = \frac{12}{35} \cdot \frac{5}{3} = \frac{4}{7}\]

\[6)\ \frac{15}{19}\ :5 = \frac{15}{19} \cdot \frac{1}{5} = \frac{3}{19}\]

\[7)\ 9\ :\frac{18}{23} = 9 \cdot \frac{23}{18} = \frac{23}{2} = 11,5\]

\[8)\ 4\frac{4^{\backslash 4}}{7} + 6\frac{1^{\backslash 7}}{4} =\]

\[= 4\frac{16}{28} + 6\frac{7}{28} = 10\frac{23}{28}\]

\[9)\ 4 - 2\frac{4}{7} = 3\frac{7}{7} - 2\frac{4}{7} = 1\frac{3}{7}\]

\[10)\ 5\frac{3^{\backslash 3}}{8} - 3\frac{5^{\backslash 4}}{6} = 5\frac{9}{24} - 3\frac{20}{24} =\]

\[= 4\frac{33}{24} - 3\frac{20}{24} = 1\frac{13}{24}\]

\[11)\ 5\frac{3}{5} \cdot 1\frac{4}{21} = \frac{28}{5} \cdot \frac{25}{21} =\]

\[= \frac{4 \cdot 5}{3} = \frac{20}{3} = 6\frac{2}{3}\]

\[12)\ 1\frac{7}{9}\ :2\frac{2}{9} = \frac{16}{9}\ :\frac{20}{9} =\]

\[= \frac{16}{9} \cdot \frac{9}{20} = \frac{4}{5}\]