Решебник самостоятельные и по алгебре 9 класс Глазков контрольные работы КР-4. Неравенства с двумя переменными и их системы. Арифметическая и геометрическая прогрессии Вариант 1

\[\boxed{Вариант\ 1.}\]

\[\boxed{\mathbf{1.}}\]

\[a_{1} = 3;\ \ a_{n + 1} = a_{n} - 2\]

\[d = - 2:\]

\[a_{5} = a_{1} + 4d = 3 + 4 \cdot ( - 2) =\]

\[= 3 - 8 = - 5.\]

\[Ответ:4).\]

\[\boxed{\mathbf{2.}}\]

\[a_{3} = 2;\ \ a_{7} = 14:\]

\[a_{3} = a_{1} + 2d \Longrightarrow a_{1} = a_{3} - 2d;\]

\[a_{7} = a_{1} + 6d \Longrightarrow a_{1} = a_{7} - 6d.\]

\[a_{3} - 2d = a_{7} - 6d\]

\[2 - 2d = 14 - 6d\]

\[- 2d + 6d = 14 - 2\]

\[4d = 12\]

\[d = 3.\]

\[a_{1} = 2 - 2d = 2 - 2 \cdot 3 = - 4.\]

\[Ответ:2).\]

\[\boxed{\mathbf{3.}}\]

\[b_{1} = 1;\ \ b_{6} = \frac{1}{243}:\]

\[b_{6} = b_{1} \cdot q^{5} \Longrightarrow q^{5} = \frac{b_{6}}{b_{1}}\]

\[q^{5} = \frac{1}{243}\ :1 = \frac{1}{243}\]

\[q = \frac{1}{3}.\]

\[S_{5} = \frac{b_{1}\left( 1 - q^{n} \right)}{1 - q} = \frac{1 \cdot \left( 1 - \left( \frac{1}{3} \right)^{5} \right)}{1 - \frac{1}{3}} =\]

\[= \frac{1 - \frac{1}{243}}{\frac{2}{3}} = \frac{242}{243} \cdot \frac{3}{2} = \frac{121}{81}.\]

\[Ответ:3).\]

\[\boxed{\mathbf{4.}}\]

\[S_{3} = 87;\ \ a_{3} + 5 = a_{1} + a_{2}.\]

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

\[a_{2} = a_{1} + d\]

\[a_{3} = a_{1} + 2d\]

\[a_{1} + a_{2} + a_{3} = a_{1} + a_{1} + d + a_{2} + 2d =\]

\[= 3a_{1} + 3d = 87\]

\[3 \cdot \left( a_{1} + d \right) = 87\]

\[a_{1} + d = 29 \Longrightarrow a_{1} = 29 - d.\]

\[a_{1} + 2d + 5 = a_{1} + a_{1} + d\]

\[- a_{1} + d + 5 = 0 \Longrightarrow a_{1} = d + 5.\]

\[29 - d = d + 5\]

\[2d = 24\]

\[d = 12.\]

\[a_{1} = 29 - 12 = 17.\]

\[a_{2} = 17 + 12 = 29.\]

\[a_{3} = 29 + 12 = 41.\]

\[Ответ:наибольшее\ число\ 41.\]

\[\boxed{\mathbf{5.}}\]

\[b_{6} - b_{4} = 8;\ \ b_{5} - b_{3} = 24.\]

\[b_{1}q^{5} - b_{1}q^{3} = b_{1}q^{3}\left( q^{2} - 1 \right) =\]

\[= q \cdot b_{1}q^{2}\left( q^{2} - 1 \right) = 8;\]

\[b_{1}q^{4} - b_{1}q^{2} = b_{1}q^{2}\left( q^{2} - 1 \right) = 24.\]

\[\left\{ \begin{matrix} q \cdot 24 = 8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ b_{1}q^{2}\left( q^{2} - 1 \right) = 24 \\ \end{matrix} \right.\ \]

\[q = \frac{8}{24} = \frac{1}{3}:\]

\[b_{1} \cdot \left( \frac{1}{3} \right)^{2}\left( \left( \frac{1}{3} \right)^{2} - 1 \right) = 24\]

\[\frac{1}{9}b_{1}\left( \frac{1}{9} - 1 \right) = 24\]

\[- \frac{8}{9}b_{1} = 24 \cdot 9\]

\[b_{1} = - \frac{24 \cdot 9 \cdot 9}{8} = - 3 \cdot 81\]

\[b_{1} = - 243.\]

\[Ответ:\ - 243;\ \frac{1}{3}.\]

\[\boxed{\mathbf{6.}}\]

\[\left\{ \begin{matrix} xy - 4 \geq 0\ \ \ \ \ \ \ \ \\ x^{2} + y - 2 \leq 0 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y \geq \frac{4}{x}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ } \\ y \leq - x^{2} + 2 \\ \end{matrix} \right.\ \]