Решебник по математике 6 класс Мерзляк ФГОС Задание 226

\[\boxed{\mathbf{226\ (226)}\mathbf{.}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \frac{12 \cdot 21}{35 \cdot 15} = \frac{4 \cdot 3}{5 \cdot 5} = \frac{12}{25}\]

\[2)\ \frac{72 \cdot 11}{33 \cdot 30} = \frac{12 \cdot 1}{3 \cdot 5} = \frac{4 \cdot 1}{5} = \frac{4}{5}\]

\[3)\ \frac{25 \cdot 17 \cdot 44}{51 \cdot 8 \cdot 75} = \frac{1 \cdot 1 \cdot 11}{3 \cdot 2 \cdot 3} = \frac{11}{18}\]

\[4)\ \frac{8 \cdot 3 + 8 \cdot 23}{3 \cdot 16} = \frac{8 \cdot (3 + 23)}{3 \cdot 16} =\]

\[= \frac{26}{3 \cdot 2} = \frac{13}{3} = 4\frac{1}{3}\]

\[5)\ \frac{17 \cdot 48}{17 \cdot 16 - 9 \cdot 16} =\]

\[= \frac{17 \cdot 48}{16 \cdot (17 - 9)} = \frac{17 \cdot 3}{1 \cdot 8} = \frac{51}{8} =\]

\[= 6\frac{3}{8}\]

\[6)\ \frac{14 \cdot 5 - 14 \cdot 3}{21 \cdot 9 + 21 \cdot 3} =\]

\[= \frac{14 \cdot (5 - 3)}{21 \cdot (9 + 3)} = \frac{2 \cdot 2}{3 \cdot 12} = \frac{1 \cdot 1}{3 \cdot 3} =\]

\[= \frac{1}{9}\]

\[\boxed{\mathbf{226}\mathbf{.}}\]

\[\angle BAC = \angle BAE - \angle CAE =\]

\[= 80{^\circ} - 56{^\circ} = 24{^\circ}.\]

\[\angle CAD = \angle BAD - \angle BAC =\]

\[= 36{^\circ} - 24{^\circ} = 12{^\circ}.\]

\[Ответ:12{^\circ}.\]