Решебник по алгебре 9 класс Мерзляк Задание 809

 

\[\boxed{\mathbf{809\ (809).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

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

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

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

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

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

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

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

\[= 2n - 4\]

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

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

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

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

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

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

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

\[Ответ:\ a_{1} = - 2;\ \ \ d = 2.\]

\[\boxed{\mathbf{809.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\frac{x^{2} - 16}{|x + 1|} \leq 0\]

\[Ответ:x \in \lbrack - 4;\ - 1) \cup ( - 1;4\rbrack.\]

\[2)\ \frac{x^{2} - 5x - 14}{|x - 8|} \geq 0\]

\[\frac{(x - 7)(x + 2)}{|x - 8|} \geq 0\]

\[Ответ:\]

\[( - \infty;\ - 2\rbrack \cup \lbrack 7;8) \cup (8;\ + \infty).\]