Решебник по алгебре 9 класс Мерзляк Задание 643

 

\[\boxed{\mathbf{643\ (643).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ Р\ (А) = \frac{10}{30} = \frac{1}{3};\ \ \ \ \]

\[(2,\ 3,\ 5,\ 7,\ 9,\ 11,\ 13,\ 17,\ 19,\ 23,\ 29).\]

\[2)\ Р\ (А) = \frac{6}{30} = \frac{1}{5};\ \ \ \]

\[(1,\ 2,\ 3,\ 6,\ 9,\ 18).\]

\[3)\ Р\ (А) = \frac{5}{30} = \frac{1}{6};\ \ \]

\[(1,\ 4,\ 9,\ 16,\ 25).\]

\[\boxed{\mathbf{643.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[S = \frac{a_{1} + a_{n}}{2} \cdot n = \frac{10 + 99}{2} \cdot 90 =\]

\[= 109 \cdot 45 = 4905 - все\]

\[\ двузначные\ числа.\]

\[Кратны\ трем:\ \ \ 12,\ 15,\ 18,\ \ldots,\]

\[\ 99,\ldots \Longrightarrow \text{\ \ }d = 3.\]

\[a_{n} = a_{1} + d(n - 1)\ \]

\[99 = 12 + 3 \cdot (n - 1)\ \]

\[99 = 12 + 3n - 3\]

\[3n = 90\]

\[n = 30.\ \ \]

\[Тогда:\]

\[S^{3} = \frac{a_{1} + a_{n}}{2} \cdot n =\]

\[= \frac{12 + 99}{2} \cdot 30 = 115 \cdot 15 =\]

\[= 1665.\]

\[Кратны\ пяти:\ \ \ \ 10\ ,15,\ \ldots,\ 95,\]

\[\ldots \Longrightarrow \text{\ \ }d = 5.\]

\[a_{n} = a_{1} + d(n - 1)\]

\[95 = 10 + 5 \cdot (n - 1)\ \]

\[95 = 10 + 5n - 5\]

\[5n = 90\]

\[n = 18.\ \ \]

\[Тогда:\]

\[S^{5} = \frac{a_{1} + a_{n}}{2} \cdot n =\]

\[= \frac{10 + 95}{2} \cdot 18 = 105 \cdot 9 = 945.\]

\[Ни\ на\ 3,\ ни\ на\ 5:\ \ 3 \cdot 5 = 15.\]

\[Делятся\ на\ 15:15,\ 30,\ 45,\ \ldots,\]

\[90 \Longrightarrow \text{\ \ }d = 15.\]

\[a_{n} = a_{1} + d(n - 1)\ \]

\[90 = 15 + 15 \cdot (n - 1)\text{\ \ }\]

\[90 = 15 + 15n - 15\]

\[15n = 90\ \]

\[\ n = 6\]

\[S^{15} = \frac{a_{1} + a_{n}}{2} \cdot n =\]

\[= \frac{15 + 90}{2} \cdot 6 = 105 \cdot 3 = 315.\]

\[Тогда\ искомая\ сумма:\]

\[S = S - S^{3} - S^{5} + S^{15} = 4905 -\]

\[- 1665 - 945 + 315 = 2610.\]

\[Ответ:2610.\]