ГДЗ по математике 6 класс Мерзляк Задание 289

\[\boxed{\mathbf{289\ (290).}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \left( x + \frac{5}{12} \right) - \frac{9}{20} = \frac{11}{15}\text{\ \ \ \ \ }\]

\[x + \frac{5}{12} = \frac{11^{\backslash 4}}{15} + \frac{9^{\backslash 3}}{20}\text{\ \ \ \ }\]

\[x + \frac{5}{12} = \frac{44 + 27}{60}\]

\[x + \frac{5}{12} = \frac{71}{60}\text{\ \ \ \ \ \ }\]

\[x = \frac{71}{60} - \frac{5^{\backslash 5}}{12}\]

\[x = \frac{71 - 25}{60} = \frac{46}{60} = \frac{23}{30}\]

\[Ответ:\ \frac{23}{30}.\]

\[2)\ \left( x - \frac{11}{30} \right) - \frac{16}{45} = \frac{2}{9}\text{\ \ \ \ \ }\]

\[x - \frac{11}{30} = \frac{2^{\backslash 5}\ }{9} + \frac{16}{45}\]

\[x - \frac{11}{30} = \frac{10}{45} + \frac{16}{45}\text{\ \ \ \ \ }\]

\[x - \frac{11}{30} = \frac{26}{45}\]

\[x = \frac{26^{\backslash 2}}{45} + \frac{11^{\backslash 3}}{30} = \frac{52 + 33}{90} =\]

\[= \frac{85}{90}\]

\[x = \frac{17}{18}\]

\[Ответ:\ \frac{17}{18}.\]

\[3)\ \left( x - \frac{7}{15} \right) + \frac{5}{8} = \frac{17}{24}\text{\ \ \ \ \ }\]

\[x - \frac{7}{15} = \frac{17}{24} - \frac{5^{\backslash 3}}{8}\text{\ \ \ \ \ }\]

\[x - \frac{7}{15} = \frac{17 - 15}{24}\]

\[x - \frac{7}{15} = \frac{2}{24}\text{\ \ \ \ \ }\]

\[x = \frac{1^{\backslash 5}}{12} + \frac{7^{\backslash 4}}{15} = \frac{5 + 28}{60} = \frac{33}{60}\]

\[x = \frac{11}{20}\]

\[Ответ:\ \frac{11}{20}.\]

\[4)\ \frac{4}{5} - \left( x + \frac{1}{60} \right) = \frac{2}{3}\text{\ \ \ }\]

\[x + \frac{1}{60} = \frac{4^{\backslash 3}}{5} - \frac{2^{\backslash 5}}{3}\text{\ \ \ \ }\]

\[x + \frac{1}{60} = \frac{12 - 10}{15}\]

\[x + \frac{1}{60} = \frac{2}{15}\text{\ \ \ }\]

\[x = \frac{2^{\backslash 4}}{15} - \frac{1}{60} = \frac{8 - 1}{60} = \frac{7}{60}\]

\[Ответ:\ \frac{7}{60}.\]

\[5)\ 4\frac{3}{4} - \left( x - 2\frac{5}{8} \right) = 3\frac{5}{6}\]

\[x - 2\frac{5}{8} = 4\frac{3^{\backslash 3}}{4} - 3\frac{5^{\backslash 2}}{6}\]

\[x - 2\frac{5}{8} = 3\frac{21}{12} - 3\frac{10}{12}\]

\[x - 2\frac{5}{8} = \frac{11}{12}\]

\[x = \frac{11^{\backslash 2}}{12} + 2\frac{5^{\backslash 3}}{8} = \frac{22}{24} + 2\frac{15}{24} =\]

\[= 2\frac{37}{24}\]

\[x = 3\frac{13}{24}\]

\[Ответ:3\frac{13}{24}.\]

\[6)\ 9\frac{9}{28} - \left( 4\frac{5}{21} - x \right) = 6\frac{2}{7}\]

\[4\frac{5}{21} - x = 9\frac{9}{28} - 6\frac{2^{\backslash 4}}{7}\]

\[4\frac{5}{21} - x = 9\frac{9}{28} - 6\frac{8}{28}\]

\[4\frac{5}{21} - x = 3\frac{1}{28}\]

\[x = 4\frac{5}{21} - 3\frac{1}{28}\]

\[x = 1\frac{5}{21} - \frac{1}{28} = \frac{26^{\backslash 4}}{21} - \frac{1^{\backslash 3}}{28} =\]

\[= \frac{104}{84} - \frac{3}{84} = \frac{101}{84}\]

\[x = 1\frac{17}{84}\]

\[Ответ:\ \ 1\frac{17}{84}.\]

\[\boxed{\mathbf{289\ (}\mathbf{с}\mathbf{)}\mathbf{.}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \frac{3}{7} + \frac{14}{19} + \frac{4}{7} + \frac{5}{19} =\]

\[= \left( \frac{3}{7} + \frac{4}{7} \right) + \left( \frac{14}{19} + \frac{5}{19} \right) =\]

\[= \frac{7}{7} + \frac{19}{19} = 1 + 1 = 2\]

\[2)\ \frac{7}{16} + \frac{11}{42} + \frac{9}{16} + \frac{17}{42} =\]

\[= \left( \frac{7}{16} + \frac{9}{16} \right) + \left( \frac{11}{42} + \frac{17}{42} \right) =\]

\[= \frac{16}{16} + \frac{28}{42} = 1 + \frac{2}{3} = 1\frac{2}{3}\]

\[3)\ \frac{5}{18} + \frac{4}{81} + \frac{7}{18} + \frac{5}{81} =\]

\[= \left( \frac{5}{18} + \frac{7}{18} \right) + \left( \frac{4}{81} + \frac{5}{81} \right) =\]

\[= \frac{12}{18} + \frac{9}{81} = \frac{2^{\backslash 3}}{3} + \frac{1}{9} = \frac{6}{9} + \frac{1}{9} =\]

\[= \frac{7}{9}\]

\[4)\ \frac{9}{40} + \frac{13}{50} + \frac{12}{50} + \frac{11}{40} =\]

\[= \frac{20}{40} + \frac{25}{50} =\]

\[= \left( \frac{9}{40} + \frac{11}{40} \right) + \left( \frac{13}{50} + \frac{12}{50} \right) =\]

\[= \frac{20}{40} + \frac{25}{50} = \frac{1}{2} + \frac{1}{2} = \frac{2}{2} = 1\]

\[5)\ 3\frac{5}{11} + 1\frac{3}{16} + 2\frac{5}{16} + 4\frac{6}{11} =\]

\[= \left( 3\frac{5}{11} + 4\frac{6}{11} \right) + \left( 1\frac{3}{16} + 2\frac{5}{16} \right) =\]

\[= 7\frac{11}{11} + 3\frac{8}{16} = 8 + 3\frac{1}{2} = 11\frac{1}{2}\]

\[6)\ 1\frac{17}{24} + 3\frac{1}{36} + 5\frac{4}{24} + 2\frac{8}{36} =\]

\[= \left( 1\frac{17}{24} + 5\frac{4}{24} \right) + \left( 3\frac{1}{36} + 2\frac{8}{36} \right) =\]

\[= 6\frac{21}{24} + 5\frac{9}{36} = 6\frac{7}{8} + 5\frac{1^{\backslash 2}}{4} =\]

\[= 6\frac{7}{8} + 5\frac{2}{8} = 11\frac{9}{8} = 12\frac{1}{8}\]