Решебник по математике 6 класс Мерзляк ФГОС Задание 288

 

\[\boxed{\mathbf{288\ (289)}\mathbf{.}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\[\boxed{\mathbf{288\ (}\mathbf{с}\mathbf{)}\mathbf{.}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \frac{2^{\backslash 2}}{9} + \frac{5^{\backslash 3}}{6} = \frac{4 + 15}{18} = \frac{19}{18} =\]

\[= 1\frac{1}{18}\ (С)\ \]

\[2)\ 1^{\backslash 17} - \frac{5}{17} = \frac{17 - 5}{17} = \frac{12}{17}\ \ \ (Т)\]

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

\[4)\ 2\frac{1^{\backslash 2}}{3} - 1\frac{1^{\backslash 3}}{2} = 2\frac{2}{6} - 1\frac{3}{6} =\]

\[= 1\frac{8}{6} - 1\frac{3}{6} = \frac{5}{6}\ \ (К)\]

\[5)\ 1\frac{1^{\backslash 4}}{7} + 2\frac{3}{28} = 1\frac{4}{28} + 2\frac{3}{28} =\]

\[= 3\frac{7}{28} = 3\frac{1}{4}\ \ (Л)\]

\[6)\ \ 5\frac{1^{\backslash 2}}{6} - 4\frac{1^{\backslash 3}}{4} = 5\frac{2}{12} - 4\frac{3}{12} =\]

\[= 4\frac{14}{12} - 4\frac{3}{12} = \frac{11}{12}\ \ (О)\]

\[7)\ \frac{1^{\backslash 3}}{4} + \frac{1}{12} + \frac{2^{\backslash 4}}{3} = \frac{3 + 1 + 8}{12} =\]

\[= \frac{12}{12} = 1\ \ (В)\]

\[\boxed{\mathbf{288}\mathbf{.}}\]

\[Сумма\ длин\ всех\ ребер:\]

\[5\ см \cdot 4 + 6\ см \cdot 4 + 10\ см \cdot 4 =\]

\[= 20\ см + 24\ см + 40\ см =\]

\[= 84\ см.\]

\[Площадь\ поверхности:\]

\[2 \cdot (5 \cdot 6 + 5 \cdot 10 + 6 \cdot 10) =\]

\[= 2 \cdot (30 + 50 + 60) = 2 \cdot 140 =\]

\[= 280\ см^{2}.\]

\[Объем:\]

\[5 \cdot 6 \cdot 10 = 300\ см^{3}.\]