ГДЗ по математике 6 класс Мерзляк Задание 291

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Год:2023
Тип:учебник
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Задание 291

\[\boxed{\mathbf{291\ (292)}\mathbf{.}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ 7\frac{7^{\backslash 4}}{9} - 4\frac{1^{\backslash 3}}{12} + 2\frac{3^{\backslash 9}}{4} =\]

\[= 7\frac{28}{36} - 4\frac{3}{36} + 2\frac{27}{36} = 5\frac{52}{36} =\]

\[= 6\frac{16}{36} = 6\frac{4}{9}\]

\[2)\ 17\frac{2^{\backslash 24}}{3} - 6\frac{1^{\backslash 2}}{36} + 4\frac{3^{\backslash 9}}{8} =\]

\[= 17\frac{48}{72} - 6\frac{2}{72} + 4\frac{27}{72} =\]

\[= 15\frac{73}{72} = 16\frac{1}{72}\]

\[3)\ 10\frac{9}{16} - \left( 3\frac{11^{\backslash 3}}{12} + 4\frac{4^{\backslash 4}}{9} \right) =\]

\[= 10\frac{9}{16} - \left( 3\frac{33}{36} + 4\frac{16}{36} \right) =\]

\[= 10\frac{9}{16} - 7\frac{49}{36} =\]

\[= 10\frac{9^{\backslash 9}}{16} - 8\frac{13^{\backslash 4}}{36} =\]

\[= 10\frac{81}{144} - 8\frac{52}{144} = 2\frac{29}{144}\]

\[4)\ \left( 20 - 7\frac{23}{36} \right) - \left( 14\frac{4^{\backslash 2}}{27} - 6\frac{1^{\backslash 3}}{18} \right) =\]

\[= \left( 19\frac{36}{36} - 7\frac{23}{36} \right) - \left( 14\frac{8}{54} - 6\frac{3}{54} \right) =\]

\[= 12\frac{13^{\backslash 3}}{36} - 8\frac{5^{\backslash 2}}{54} =\]

\[= 12\frac{39}{108} - 8\frac{10}{108} = 4\frac{29}{108}\]

\[\boxed{\mathbf{291\ (}\mathbf{с}\mathbf{)}\mathbf{.}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \left( x + \frac{4}{21} \right) - \frac{4}{15} = \frac{16}{35}\text{\ \ \ \ }\]

\[x + \frac{4}{21} = \frac{16^{\backslash 3}}{35} + \frac{4^{\backslash 7}}{15}\text{\ \ \ \ \ }\]

\[x + \frac{4}{21} = \frac{48}{105} + \frac{28}{105}\]

\[x + \frac{4}{21} = \frac{76}{105}\text{\ \ \ \ }\]

\[x = \frac{76}{105} - \frac{4^{\backslash 5}}{21} = \frac{76 - 20}{105} = \frac{56}{105}\]

\[x = \frac{8}{15}\]

\[Ответ:\ \frac{8}{15}.\]

\[2)\ \left( x - \frac{8}{19} \right) - \frac{4}{57} = \frac{2}{3}\text{\ \ \ \ \ }\]

\[x - \frac{8}{19} = \frac{2^{\backslash 19}}{3} + \frac{4}{57}\]

\[x - \frac{8}{19} = \frac{38}{57} + \frac{4}{57}\text{\ \ \ \ \ \ }\]

\[x - \frac{8}{19} = \frac{42}{57}\]

\[x = \frac{42}{57} + \frac{8^{\backslash 3}}{19} = \frac{42 + 24}{57} = \frac{66}{57} =\]

\[= \frac{22}{19}\]

\[x = 1\frac{3}{19}.\]

\[3)\ \left( x - \frac{8}{9} \right) + \frac{3}{8} = \frac{19}{36}\text{\ \ \ \ \ }\]

\[x - \frac{8}{9} = \frac{19^{\backslash 2}}{36} - \frac{3^{\backslash 9}}{8}\text{\ \ \ \ \ }\]

\[x - \frac{8}{9} = \frac{38 - 27}{72}\]

\[x - \frac{8}{9} = \frac{11}{72}\text{\ \ \ \ }\]

\[\ x = \frac{11}{72} + \frac{8^{\backslash 8}}{9} = \frac{11 + 64}{72} = \frac{75}{72}\]

\[x = 1\frac{3}{72} = 1\frac{1}{24}\]

\[Ответ:1\frac{1}{24}.\]

\[4)\ 3\frac{1}{6} - \left( x + 1\frac{1}{12} \right) = \frac{1}{4}\text{\ \ \ }\]

\[x + 1\frac{1}{12} = 3\frac{1^{\backslash 2}}{6} - \frac{1^{\backslash 3}}{4}\text{\ \ \ \ \ }\]

\[x + 1\frac{1}{12} = 3\frac{2}{12} - \frac{3}{12}\]

\[x + 1\frac{1}{12} = 2\frac{14}{12} - \frac{3}{12}\text{\ \ \ \ }\]

\[x + 1\frac{1}{12} = 2\frac{11}{12}\text{\ \ \ \ }\]

\[x = 2\frac{11}{12} - 1\frac{1}{12}\]

\[x = 1\frac{10}{12} = 1\frac{5}{6}\]

\[Ответ:1\frac{5}{6}.\]

\[5)\ 6\frac{5}{27} - \left( x - 1\frac{2}{9} \right) = 3\frac{20}{81}\text{\ \ \ \ }\]

\[x - 1\frac{2}{9} = 6\frac{5^{\backslash 3}}{27} - 3\frac{20}{81}\]

\[x - 1\frac{2}{9} = 6\frac{15}{81} - 3\frac{20}{81}\text{\ \ \ \ \ \ \ }\]

\[x - 1\frac{2}{9} = 5\frac{96}{81} - 3\frac{20}{81}\text{\ \ \ }\]

\[x - 1\frac{2}{9} = 2\frac{76}{81}\]

\[x = 2\frac{76}{81} + 1\frac{2^{\backslash 9}}{9} = 2\frac{76}{81} + 1\frac{18}{81}\]

\[x = 3\frac{94}{81} = 4\frac{13}{81}\]

\[Ответ:4\frac{13}{81}.\]

\[6)\ 3\frac{5}{36} - \left( 1\frac{4}{9} - x \right) = \frac{17}{18}\text{\ \ \ \ \ }\]

\[1\frac{4}{9} - x = 3\frac{5}{36} - 1\frac{17^{\backslash 2}}{18}\]

\[1\frac{4}{9} - x = 3\frac{5}{36} - 1\frac{34}{36}\text{\ \ \ \ }\]

\[1\frac{4}{9} - x = 2\frac{41}{36} - 1\frac{34}{36}\]

\[1\frac{4}{9} - x = 1\frac{7}{36}\text{\ \ \ }\]

\[x = 1\frac{4^{\backslash 4}}{9} - 1\frac{7}{36} = 1\frac{16}{36} - 1\frac{7}{36}\]

\[x = \frac{9}{36} = \frac{1}{4}\]

\[Ответ:\ \frac{1}{4}.\]

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