Решебник по алгебре 7 класс Мерзляк ФГОС Задание 8

 

\[\boxed{\text{8\ (8).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ 2x - 3;\text{\ \ }\]

\[при\ x = 4;\ \ 0;\ - 3:\]

\[2 \cdot 4 - 3 = 8 - 3 = 5\]

\[2 \cdot 0 - 3 = - 3\]

\[2 \cdot ( - 3) - 3 = - 6 - 3 = - 9.\]

\[2)\ \frac{1}{3}a + \frac{1}{4}b;\]

\[при\ \ a = - 6;b = 16:\]

\[\frac{1}{3} \cdot ( - 6) + \frac{1}{4} \cdot 16 = - \frac{6}{3} + \frac{16}{4} =\]

\[= - 2 + 4 = 2.\]

\[3)\ 3m - 5n + 3k;\]

\[при\ \ m = - 7;n = 1,4;k = - 0,1:\]

\[3 \cdot ( - 7) - 5 \cdot 1,4 + 3 \cdot ( - 0,1) =\]

\[= - 21 - 7 - 0,3 = - 28,3.\]

\[\boxed{\text{8.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ \frac{1^{\backslash 2}}{3} + \frac{5}{6} = \frac{2}{6} + \frac{5}{6} = \frac{7}{6} = 1\frac{1}{6}\]

\[2)\ \frac{3^{\backslash 9}}{7} - \frac{2^{\backslash 7}}{9} = \frac{27}{63} - \frac{14}{63} = \frac{13}{63}\]

\[3)\ \frac{7}{16} \cdot \frac{8}{35} = \frac{7 \cdot 8}{16 \cdot 35} = \frac{1 \cdot 1}{2 \cdot 5} =\]

\[= \frac{1}{10}\]

\[4)\ \frac{4}{9} \cdot 18 = \frac{4 \cdot 18}{9} = 4 \cdot 2 = 8\]

\[5)\ \frac{46}{75}\ :\frac{23}{45} = \frac{46 \cdot 45}{75 \cdot 23} = \frac{2 \cdot 3}{5 \cdot 1} =\]

\[= \frac{6}{5} = 1\frac{1}{5}\]

\[6)\ \frac{2}{3}\ :4 = \frac{2}{3 \cdot 4} = \frac{1}{6}\]

\[7)\ 10\ :\frac{5}{11} = \frac{10 \cdot 11}{5} = 2 \cdot 11 =\]

\[= 22\]

\[8)\ 2\frac{3^{\backslash 3}}{8} + 4\frac{1^{\backslash 4}}{6} = 2\frac{9}{24} + 4\frac{4}{24} =\]

\[= 6\frac{13}{24}\]

\[9)\ 6 - 1\frac{3}{5} = 5\frac{5}{5} - 1\frac{3}{5} = 4\frac{2}{5}\]

\[10)\ 4\frac{2^{\backslash 9}}{7} - 1\frac{4^{\backslash 7}}{9} =\]

\[= 4\frac{18}{63} - 1\frac{28}{63} = 3\frac{81}{63} - 1\frac{28}{63} =\]

\[= 2\frac{53}{63}\]

\[11)\ 8\frac{3}{4} \cdot 1\frac{3}{14} =\]

\[= \frac{35}{4} \cdot \frac{17}{14} = \frac{5 \cdot 17}{4 \cdot 2} = \frac{85}{8} = 10\frac{5}{8}\]

\[12)\ 1\frac{3}{5}\ :5\frac{1}{3} = \frac{8}{5}\ :\frac{16}{3} = \frac{8 \cdot 3}{5 \cdot 16} =\]

\[= \frac{1 \cdot 3}{5 \cdot 2} = \frac{3}{10}\]