Решебник по математике 6 класс Мерзляк ФГОС Задание 290

 

\[\boxed{\mathbf{290\ (291)}\mathbf{.}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

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

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

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

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

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

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

\[Ответ:\ \frac{8}{15}.\]

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

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

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

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

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

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

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

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

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

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

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

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

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

\[Ответ:1\frac{1}{24}.\]

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

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

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

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

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

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

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

\[Ответ:1\frac{5}{6}.\]

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

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

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

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

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

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

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

\[Ответ:4\frac{13}{81}.\]

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

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

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

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

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

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

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

\[Ответ:\ \frac{1}{4}.\]

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

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

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

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

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

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

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

\[Ответ:\ \frac{23}{30}.\]

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

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

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

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

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

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

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

\[Ответ:\ \frac{17}{18}.\]

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

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

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

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

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

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

\[Ответ:\ \frac{11}{20}.\]

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

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

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

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

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

\[Ответ:\ \frac{7}{60}.\]

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

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

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

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

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

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

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

\[Ответ:3\frac{13}{24}.\]

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

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

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

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

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

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

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

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

\[Ответ:\ \ 1\frac{17}{84}.\]

\[\boxed{\mathbf{290}\mathbf{.}}\]

\[1)\ ABC;\]

\[2)\ M;\]

\[3)\ MAB;MBC;MAC;\]

\[4)\ MA;MB;MC;\]

\[5)\ AB;BC;AC.\]