Решебник по алгебре 9 класс Мерзляк Задание 642

 

\[\boxed{\mathbf{642\ (642).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ Р\ (12) = \frac{1}{23}\]

\[2)\ Р\ (24) = 0\]

\[3)\ Р\ (А) = \frac{11}{23};\ \]

\[(2,\ 4,\ 6,\ 8,\ 10,\ 12,\ 14,\ 16,\ 18,\ 20\ ,22)\]

\[4)\ Р\ (А) = \frac{12}{23};\ (23 - 11 = 12)\]

\[5)\ Р\ (А) = \frac{7}{23};\ \ \ \]

\[(3,\ 6,\ 9,\ 12,\ 18,\ 21,\ 15)\]

\[6)\ Р\ (А) = \frac{3}{23};\ \ \ (7,\ 14,\ 21)\]

\[7)\ Р\ (А) = \frac{14}{23};\ \ \ (23 - 9 = 14)\]

\[8)\ Р\ (А) = \frac{9}{23};\ \ \ \]

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

\[9)\ Р\ (А) = \frac{2}{23};\ \ (9,\ 19)\]

\[10)\ Р\ (А) = \frac{12}{23};\ \ \ \]

\[(1;\ 10 - 19;\ 21)\]

\[11)\ Р\ (А) = \frac{21}{23};\ \ \]

\[(5;\ \ 15;\ \ 23 - 2 = 21)\]

\[12)\ Р\ (А) = \frac{3}{23};\ \ \ (5;14;23)\]

\[13)\ Р\ (А) = \frac{2}{23};\ \ \ (12;19)\]

\[14)\ Р\ (А) = \frac{11}{23};\ \ \ \]

\[(23 - 12 = 11)\]

\[\boxed{\mathbf{642.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[S_{n} = n^{2} - 3n\]

\[a_{n} = S_{n} - S_{n - 1} = n^{2} - 3n -\]

\[- \left( (n - 1)^{2} - 3 \cdot (n - 1) \right) =\]

\[= n^{2} - 3n -\]

\[- \left( n^{2} - 2n + 1 - 3n + 3 \right) = n^{2} -\]

\[- 3n - \left( n^{2} - 5n + 4 \right) =\]

\[= n^{2} - 3n - n^{2} + 5n - 4 =\]

\[= 2n - 4\]

\[a_{n + 1} - a_{n} = 2 \cdot (n + 1) - 4 -\]

\[- (2n - 4) = 2n + 2 - 4 -\]

\[- 2n + 4 = 2,\]

\[тогда\ при\ любом\ n \in N \Longrightarrow\]

\[\Longrightarrow \ a_{n + 1} = a_{n} + 2 \Longrightarrow d = 2.\ \]

\[S_{n} = n^{2} - 3n \Longrightarrow \text{\ \ \ }\]

\[{\Longrightarrow a}_{1} = S_{1} = 1 - 3 \cdot 1 = - 2.\]

\[Ответ:\ a_{1} = - 2;\ \ \ d = 2.\]