ГДЗ самостоятельные и по алгебре 9 класс Глазков контрольные работы КР-4. Неравенства с двумя переменными и их системы. Арифметическая и геометрическая прогрессии Вариант 3

\[\boxed{Вариант\ 3.}\]

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

\[a_{1} = 4;\ \ a_{n + 1} = a_{n} - 3.\]

\[d = - 3:\]

\[a_{6} = a_{1} + 5d = 4 + 5 \cdot ( - 3) = - 11.\]

\[Ответ:3.\]

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

\[a_{6} = 6;\ \ a_{9} = 15:\]

\[a_{6} = a_{1} + 5d \Longrightarrow a_{1} = a_{6} - 5d;\]

\[a_{9} = a_{1} + 8d \Longrightarrow a_{1} = a_{9} - 8d.\]

\[a_{6} - 5d = a_{9} - 8d\]

\[- 5d + 8d = a_{9} - a_{6}\]

\[3d = 15 - 6\]

\[3d = 9\]

\[d = 3.\]

\[a_{1} = a_{6} - 5d = 6 - 5 \cdot 3 = 6 - 15 = - 9.\]

\[Ответ:1).\]

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

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

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

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

\[q = 2.\]

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

\[= 64 - 1 = 63.\]

\[Ответ:3).\]

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

\[a_{1} + a_{2} = 25;\ \ a_{2} + a_{3} = 39.\]

\[1)\ a_{1} + a_{1} + d = 25\]

\[2a_{1} + d = 25\]

\[2a_{1} = 25 - d.\]

\[2)\ a_{1} + d + a_{1} + 2d = 39\]

\[2a_{1} + 3d = 39\]

\[2a_{1} = 39 - 3d.\]

\[3)\ 25 - d = 39 - 3d\]

\[- d + 3d = 39 - 25\]

\[2d = 14\]

\[d = 7.\]

\[4)\ 2a_{1} = 25 - d = 25 - 7\]

\[2a_{1} = 18\]

\[a_{1} = 9.\]

\[a_{2} = 9 + 7 = 16;\]

\[a_{3} = 16 + 7 = 23.\]

\[Наибольшее\ число:23.\]

\[Ответ:23.\]

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

\[b_{4} - b_{2} = 36:\]

\[b_{1}q^{3} - b_{1}q = b_{1}q\left( q^{2} - 1 \right) = 36.\ \ \]

\[b_{5} - b_{3} = 18.\]

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

\[= q \cdot b_{1}q^{2}\left( q^{2} - 1 \right) = 18.\]

\[q \cdot 36 = 18\]

\[q = \frac{18}{36} = \frac{1}{2} = 0,5.\]

\[b_{1} \cdot \frac{1}{2} \cdot \left( \left( \frac{1}{2} \right)^{2} - 1 \right) = 36\]

\[\frac{1}{2}b_{1} \cdot \left( \frac{1}{4} - 1 \right) = 36\]

\[b_{1} \cdot \left( - \frac{3}{4} \right) = 72\]

\[b_{1} = - 72 \cdot \frac{4}{3} = - 24 \cdot 4 = - 96.\]

\[Ответ:\ - 96;\ \ 0,5.\]

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

\[\left\{ \begin{matrix} x^{2} - y - 1 \leq 0 \\ xy - 2 \geq 0\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y \geq x^{2} - 1 \\ y \geq \frac{2}{x}\text{\ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]