Решебник по математике 6 класс Мерзляк ФГОС Задание 463

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Год:2020-2021-2022-2023
Тип:учебник
Серия:Алгоритм успеха

Задание 463

Выбери издание
Математика 6 класс Мерзляк ФГОС, Полонский, Якир. Новое издание Вентана-Граф
 
фгос Математика 6 класс Мерзляк ФГОС, Полонский Просвещение
Издание 1
Математика 6 класс Мерзляк ФГОС, Полонский, Якир. Новое издание Вентана-Граф

\[\boxed{\mathbf{463\ (463).}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ 5\frac{11}{14}x - \frac{8}{15} = \frac{5}{21}\]

\[5\frac{11}{14}x = \frac{5^{\backslash 5}}{21} + \frac{8^{\backslash 7}}{15}\]

\[\frac{81}{14}x = \frac{25 + 56}{105}\]

\[\frac{81}{14}x = \frac{81}{105}\]

\[x = \frac{81}{105}\ :\frac{81}{14}\]

\[x = \frac{81}{105} \cdot \frac{14}{81}\]

\[x = \frac{2}{15}\]

\[Ответ:\ \frac{2}{15}.\]

\[2)\ 7\frac{3}{10} + \frac{25}{28}x = 8\frac{13}{35}\]

\[\frac{25}{28}x = 8\frac{13^{\backslash 2}}{35} - 7\frac{3^{\backslash 7}}{10}\]

\[\frac{25}{28}x = 1\ \frac{26 - 21}{70}\]

\[\frac{25}{28}x = 1\frac{5}{70}\]

\[\frac{25}{28}x = \frac{75}{70}\]

\[x = \frac{75}{70}\ :\frac{25}{28}\]

\[x = \frac{75}{70} \cdot \frac{28}{25}\]

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

\[Ответ:\ \ 1\frac{1}{5}.\]

\[3)\ 3\frac{1}{3} - 1\frac{1}{20}x = 1\frac{14}{15}\]

\[\frac{21}{20}x = \frac{10^{\backslash 5}}{3} - \frac{29}{15}\]

\[\frac{21}{20}x = \frac{50 - 29}{15}\]

\[\frac{21}{20}x = \frac{21}{15}\]

\[x = \frac{21}{15}\ :\frac{21}{20}\]

\[x = \frac{21}{15} \cdot \frac{20}{21}\]

\[x = \frac{4}{3} = 1\frac{1}{3}\]

\[Ответ:1\frac{1}{3}.\]

\[4)\frac{3^{\backslash 3}}{8}x + \frac{7^{\backslash 2}}{12}x - \frac{5^{\backslash 4}}{6}x = \frac{9}{32}\]

\[\frac{9 + 14 - 20}{24}x = \frac{9}{32}\]

\[\frac{3}{24}x = \frac{9}{32}\]

\[x = \frac{9}{32}\ :\frac{3}{24}\]

\[x = \frac{9}{32} \cdot \frac{24}{3}\]

\[x = \frac{9}{4} = 2\frac{1}{4}\]

\[Ответ:2\frac{1}{4}.\]

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

\[\frac{7}{3}\ :x = 1\frac{5}{9} + 1\frac{1}{6}\]

\[\frac{7}{3}\ :x = \frac{14^{\backslash 2}}{9} + \frac{7^{\backslash 3}}{6}\]

\[\frac{7}{3}\ :x = \frac{28 + 21}{18}\]

\[\frac{7}{3}\ :x = \frac{49}{18}\]

\[x = \frac{7}{3}\ :\frac{49}{18}\]

\[x = \frac{7}{3} \cdot \frac{18}{49}\]

\[x = \frac{6}{7}\]

\[Ответ:\ \frac{6}{7}.\]

\[6)\ 2\frac{1}{3}\ :\left( x - 1\frac{1}{6} \right) = 1\frac{5}{9}\]

\[\frac{7}{3}\ :\left( x - \frac{7}{6} \right) = \frac{14}{9}\]

\[x - \frac{7}{6} = \frac{7}{3}\ :\frac{14}{9}\]

\[x - \frac{7}{6} = \frac{7}{3} \cdot \frac{9}{14}\]

\[x = \frac{3^{\backslash 3}}{2} + \frac{7}{6}\]

\[x = \frac{9 + 7}{6} = \frac{16}{6}\]

\[x = \frac{8}{3} = 2\frac{2}{3}\]

\[Ответ:\ 2\frac{2}{3}.\]

\[7)\ 27\ :\left( 31\frac{3}{7} - 2\frac{11}{14}x \right) = 1\frac{1}{8}\]

\[31\frac{3}{7} - 2\frac{11}{14}x = 27\ :\frac{9}{8}\]

\[\frac{39}{14}x = 31\frac{3}{7} - 24\]

\[\frac{39}{14}x = 7\frac{3}{7}\]

\[x = \frac{52}{7}\ :\frac{39}{14}\]

\[x = 2\frac{2}{3}\]

\[Ответ:2\frac{2}{3}.\]

\[8)\ 48\ :\left( 3\frac{4}{5}x - 25 \right) = 1\frac{1}{2}\]

\[3\frac{4}{5}x - 25 = 48\ :\frac{3}{2}\]

\[\frac{19}{5}x - 25 = 48 \cdot \frac{2}{3}\]

\[\frac{19}{5}x = 32 + 25\]

\[\frac{19}{5}x = 57\]

\[x = 57\ :\frac{19}{5}\]

\[x = 57 \cdot \frac{5}{19}\]

\[Ответ:15.\]

Издание 2
фгос Математика 6 класс Мерзляк ФГОС, Полонский Просвещение

\[\boxed{\mathbf{463}.}\]

\[1)\ 17,5\ :7 = 175\ :70 = 2,5\]

\[2)\ 20,4\ :24 = 204\ :240 = 0,85\]

\[3)\ 453,2\ :22 =\]

\[= 4532\ :220 = 20,6\]

\[4)\ 5,49\ :6 = 549\ :600 = 0,915\]

\[5)\ 0,1242\ :6,9 =\]

\[= 1242\ :69\ 000 = 0,018\]

\[6)\ 0,02592\ :0,048 =\]

\[= 2592\ :4800 = 0,54\]

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