Решебник по математике 6 класс Мерзляк ФГОС Задание 361

\[\boxed{\mathbf{361\ (361).}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

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

\[= \frac{6 + 4 - 3}{12}a = \frac{7}{12}a\]

\[a = 1\frac{5}{7} = \frac{12}{7}:\]

\[\frac{7}{12}a = \frac{7 \cdot 12}{12 \cdot 7} = 1.\]

\[2)\ \frac{4^{\backslash 3}}{7}b + \frac{5}{21}b - \frac{2^{\backslash 7}}{3}b =\]

\[= \frac{12 + 5 - 14}{21}b = \frac{3}{21}b = \frac{1}{7}b\]

\[b = 2\frac{1}{3} = \frac{7}{3}:\]

\[\frac{1}{7}b = \frac{1 \cdot 7}{7 \cdot 3} = \frac{1}{3}.\]

\[3)\ 1\frac{5^{\backslash 3}}{12}m + 2\frac{7^{\backslash 2}}{18}m - 1\frac{2^{\backslash 4}}{9}m =\]

\[= \left( 1\frac{15}{36} + 2\frac{14}{36} - 1\frac{8}{36} \right)m =\]

\[= 2\frac{21}{36}m = 2\frac{7}{12}m\]

\[m = 1\frac{17}{31} = \frac{48}{31}:\]

\[\boxed{\mathbf{361}\mathbf{.}}\]

\[1)\ 2\frac{6}{7}\ :\left( \frac{5^{\backslash 7}}{6} - \frac{9^{\backslash 3}}{14} \right) =\]

\[= \frac{20}{7}\ :\frac{35 - 27}{42} = \frac{20}{7}\ :\frac{8}{42} =\]

\[= \frac{20}{7} \cdot \frac{42}{8} = \frac{5 \cdot 6}{1 \cdot 2} = 5 \cdot 3 = 15\]

\[2)\ 2\frac{6}{7}\ :\frac{5}{6} - \frac{9}{14} = \frac{20}{7} \cdot \frac{6}{5} - \frac{9}{14} =\]

\[= \frac{24^{\backslash 2}}{7} - \frac{9}{14} = \frac{48 - 9}{14} = \frac{39}{14} =\]

\[= 2\frac{11}{14}\]