Решебник по алгебре 9 класс Мерзляк Задание 779

 

\[\boxed{\mathbf{779\ (779).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[S_{5} = 100,\ \ n = 5,\]

\[\text{\ \ }тогда\ 7 \cdot \left( a_{1} + a_{2} \right) =\]

\[= a_{3} + a_{4} + a_{5},\]

\[S_{5} = \frac{a_{1} + a_{5}}{2} \cdot 5;\ \ \ a_{1} + a_{5} = 40\ \]

\[Составим\ систему:\]

\[\left\{ \begin{matrix} a_{1} + a_{5} = 40\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 7 \cdot \left( a_{1} + a_{2} \right) = a_{3} + a_{4} + a_{5} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 2a_{1} + 4d = 40\ \ \ \ \ \ \ |\ :2 \\ 14a_{1} + 7d = 3a_{1} + 9d \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} a_{1} + 2d = 20 \\ 11a_{1} - 2d = 0 \\ \end{matrix} \right.\ + \ \ \ \ \ \ \ \]

\[\left\{ \begin{matrix} 12a_{1} = 20 \\ d = \frac{11}{2}a_{1} \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\left\{ \begin{matrix} a_{1} = 1\frac{2}{3}\text{\ \ \ } \\ d = \frac{11 \cdot 5}{2 \cdot 3} \\ \end{matrix}\text{\ \ } \right.\ \text{\ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} a_{1} = 1\frac{2}{3} \\ d = 9\frac{1}{6}\ \\ \end{matrix} \right.\ \]

\[a_{2} = 1\frac{2}{3} + 9\frac{1}{6} = 10\frac{5}{6};\ \]

\[\ a_{3} = 10\frac{5}{6} + 9\frac{1}{6} = 20\]

\[a_{4} = 20 + 9\frac{1}{6} = 29\frac{1}{6};\ \]

\[\ a_{5} = 29\frac{1}{6} + 9\frac{1}{6} = 38\frac{1}{3}\]

\[Ответ:1\frac{2}{3};\ 10\frac{5}{6};\ \ 20;\]

\[\ \ 29\frac{1}{6};\ \ 38\frac{1}{3}.\]

\[\boxed{\mathbf{779.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[f(x) = 3x + 12\]

\[1)\ 3x + 12 > 0,\ \ 3x > - 12,\]

\[\ \ x > - 4\]

\[Ответ:( - 4;\ + \infty).\]

\[2)\ 3x + 12 < 0,\ \ 3x < - 12,\]

\[\ \ x < - 4\]

\[Ответ:( - \infty;\ - 4).\]

\[3) - 4 \leq 3x + 12 \leq 7\]

\[- 4 - 12 \leq 3x \leq 7 - 12\]

\[- 16 \leq 3x \leq - 5\]

\[- \frac{16}{3} \leq x \leq - \frac{5}{3}\]