ГДЗ самостоятельные и по алгебре 9 класс Глазков контрольные работы КР-4. Неравенства с двумя переменными и их системы. Арифметическая и геометрическая прогрессии Вариант 2

\[\boxed{Вариант\ 2.}\]

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

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

\[d = 2:\]

\[a_{6} = a_{1} + 5d = - 3 + 2 \cdot 5 = 7.\]

\[Ответ:2).\]

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

\[a_{8} = 5;\ \ a_{10} = 13:\]

\[a_{8} = a_{1} + 7d \Longrightarrow a_{1} = a_{8} - 7d;\]

\[a_{10} = a_{1} + 9d \Longrightarrow a_{1} = a_{10} - 9d.\]

\[a_{8} - 7d = a_{10} - 9d\]

\[5 - 7d = 13 - 9d\]

\[- 7d + 9d = 13 - 5\]

\[2d = 8\]

\[d = 4.\]

\[a_{1} = 5 - 7 \cdot 4 = 5 - 28 = - 23.\]

\[Ответ:1).\]

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

\[b_{1} = 1;\ \ b_{4} = \frac{1}{8}:\]

\[b_{4} = b_{1} \cdot q^{3} \Longrightarrow q^{3} = \frac{b_{4}}{b_{1}};\]

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

\[q = \frac{1}{2}.\]

\[S_{6} = \frac{b_{1}\left( 1 - q^{n} \right)}{1 - q} = \frac{1 \cdot \left( 1 - \left( \frac{1}{2} \right)^{6} \right)}{1 - \frac{1}{2}} =\]

\[= \frac{1 - \frac{1}{64}}{\frac{1}{2}} = \frac{63}{64}\ :\frac{1}{2} = \frac{63 \cdot 2}{64} = \frac{63}{32}.\]

\[Ответ:4).\]

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

\[a_{1} + a_{2} = 132;\ \ \frac{a_{3}}{a_{1}} = 3:\]

\[1)\ a_{1} + a_{1} + d = 132\]

\[2a_{1} + d = 132\]

\[2a_{1} = 132 - d.\]

\[2)\ a_{3} = 3a_{1}\]

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

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

\[3)\ 132 - d = 2d\]

\[3d = 132\]

\[d = 44.\]

\[4)\ 2a_{1} = 2d = 88\]

\[a_{1} = 44.\]

\[a_{2} = 44 + 44 = 88.\]

\[a_{3} = 88 + 44 = 132.\]

\[Наименьшее\ число:44.\]

\[Ответ:44.\]

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

\[b_{6} - b_{4} = 144:\]

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

\[= q \cdot b_{1}q^{2}\left( q^{2} - 1 \right) = 144.\]

\[b_{5} - b_{3} = 48:\]

\[b_{1}q^{4} - b_{1}q^{2} = b_{1}q^{2}\left( q^{2} - 1 \right) = 48.\]

\[q \cdot 48 = 144\]

\[q = \frac{144}{48} = 3.\]

\[b_{1} \cdot (3)^{2} \cdot \left( 3^{2} - 1 \right) = 48\]

\[9b_{1} \cdot 8 = 48\]

\[9b_{1} = 6\]

\[b_{1} = \frac{6}{9}\]

\[b_{1} = \frac{2}{3}.\]

\[Ответ:\ \frac{2}{3};\ \ 3.\]

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

\[\left\{ \begin{matrix} y \geq (x - 1)^{2}\text{\ \ \ \ } \\ 2x - y + 5 \geq 0 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y \geq (x - 1)^{2} \\ y \leq 2x + 5\ \ \ \\ \end{matrix} \right.\ \]