Решебник по математике 6 класс Мерзляк ФГОС Задание 292

 

\[\boxed{\mathbf{292\ (293)}\mathbf{.}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ 5\frac{5^{\backslash 6}}{9} + 3\frac{1^{\backslash 9}}{6} - 6\frac{4^{\backslash 2}}{27} =\]

\[= 5\frac{30}{54} + 3\frac{9}{54} - 6\frac{8}{54} = 2\frac{31}{54}\]

\[2)\ 1\frac{5^{\backslash 4}}{7} + 3\frac{11^{\backslash 2}}{14} - 2\frac{1^{\backslash 7}}{4} =\]

\[= 1\frac{20}{28} + 3\frac{22}{28} - 2\frac{7}{28} = 2\frac{35}{28} =\]

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

\[3)\ 12\frac{13^{\backslash 2}}{48} - \left( 9\frac{17^{\backslash 3}}{32} - 4\frac{5^{\backslash 4}}{24} \right) =\]

\[= 12\frac{26}{96} - \left( 9\frac{51}{96} - 4\ \frac{20}{96} \right) =\]

\[= 12\frac{26}{96} - 5\frac{31}{96} =\]

\[= 11\frac{\ 122}{96} - 5\frac{31}{96} = 6\frac{91}{96}\]

\[4)\ \left( 18 - 10\frac{18}{35} \right) - \left( 3\frac{9^{\backslash 5}}{28} + 2\frac{3^{\backslash 7}}{20} \right) =\]

\[= \left( 17\frac{35}{35} - 10\frac{18}{35} \right) - \left( 3\frac{45}{140} + 2\frac{21}{140} \right) =\]

\[= 7\frac{17}{35} - 5\frac{66}{140} =\]

\[= 7\frac{17^{\backslash 2}}{35} - 5\frac{33}{70} = 7\frac{34}{70} - 5\frac{33}{70} =\]

\[= 2\frac{1}{70}\]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\[\boxed{\mathbf{292}\mathbf{.}}\]

\[Объем\ одного\ кубика = 1\ см^{3}.\]

\[\textbf{а)}\ 20\ кубиков:\]

\[V = 20 \cdot 1 = 20\ см^{3}.\]

\[\textbf{б)}\ 21\ кубик:\]

\[V = 21 \cdot 1 = 21\ см^{3}.\]