Решебник по математике 6 класс Мерзляк ФГОС Задание 278

\[\boxed{\mathbf{278\ (277)}\mathbf{.}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ 8\frac{9}{14} - 3\frac{3^{\backslash 2}}{7} = 8\frac{9}{14} - 3\frac{6}{14} =\]

\[= 5\frac{3}{14}\]

\[2)\ 7\frac{5^{\backslash 2}}{12} - 3\frac{7}{24} = 7\frac{10}{24} - 3\frac{7}{24} =\]

\[= 4\frac{3}{24} = 4\frac{1}{8}\]

\[3)\ 12\frac{11^{\backslash 3}}{12} - 5\frac{13^{\backslash 2}}{18} =\]

\[= 12\frac{33}{36} - 5\frac{26}{36} = 7\frac{7}{36}\]

\[\boxed{\mathbf{278\ (}\mathbf{с}\mathbf{)}\mathbf{.}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ 1^{\backslash 29} - \frac{17}{29} = \frac{29 - 17}{29} = \frac{12}{29}\]

\[2)\ 4 - \frac{7}{19} = 3\frac{19}{19} - \frac{7}{19} = 3\frac{2}{19}\]

\[3)\ 13 - 7\frac{3}{16} = 12\frac{16}{16} - 7\frac{3}{16} =\]

\[= 5\frac{13}{16}\]

\[4)\ 15 - 12\frac{16}{17} = 14\frac{17}{17} - 12\frac{16}{17} =\]

\[= 2\frac{1}{17}\]

\[\boxed{\mathbf{278}\mathbf{.}}\]

\[\angle ACB = 90{^\circ};\]

\[\angle ADB = 90{^\circ}.\]