ГДЗ по алгебре 9 класс Мерзляк Проверь себя №6

\[\boxed{\mathbf{Задание}\mathbf{\ }\mathbf{№}\mathbf{6.}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[\mathbf{№1.}\]

\[\mathbf{Ответ:Б).}\]

\[\mathbf{№2.}\]

\[\mathbf{Ответ:В).}\]

\[\mathbf{№3.}\]

\[a_{6} = a_{1} + 5d = 12 +\]

\[+ 5 \cdot 0,4 = 14\]

\[\mathbf{Ответ:Б).}\]

\[\mathbf{№4.}\]

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

\[d = a_{2} - a_{1} = 5 + 7 = 12\]

\[\mathbf{Ответ:Г).}\]

\[\mathbf{№5.}\]

\[S_{10} = \frac{2a_{1} + 9d}{2} \cdot 10 =\]

\[= \left( 2a_{1} + 9d \right) \cdot 5 =\]

\[= ( - 32 + 27) \cdot 5 = - 25\]

\[\mathbf{Ответ:Г).}\]

\[\mathbf{№6.}\]

\[b_{n} = b_{1}q^{3} = - \frac{1}{8} \cdot ( - 2)^{3} =\]

\[= - \frac{1}{8} \cdot ( - 8) = 1\]

\[\mathbf{Ответ:В).}\]

\[\mathbf{№7.}\]

\[b_{2} = b_{1}q,\ \ q = \frac{b_{2}}{b_{1}} = \frac{9}{36} = \frac{1}{4}\]

\[\mathbf{Ответ:А).}\]

\[\mathbf{№8.}\]

\[S_{n} = \frac{2 \cdot \left( 3^{4} - 1 \right)}{3 - 1} =\]

\[= \frac{2 \cdot (81 - 1)}{2} = 80\]

\[Ответ:Б).\]

\[\mathbf{№9.}\]

\[a_{n} = - 4n + 13\ \]

\[a_{1} = - 4 + 13 = 9\]

\[a = - 4 \cdot 2 + 13 = 5\]

\[a_{15} = - 4 \cdot 15 + 13 = - 47\]

\[S_{15} = \frac{9 - 47}{2} \cdot 15 =\]

\[= - 19 \cdot 15 = - 285\]

\[Ответ:Б).\]

\[\mathbf{№10.}\]

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

\[6,2 = 0,2 + 0,4 \cdot (n - 1)\ \]

\[\ 0,4 \cdot (n - 1) = 6\ \ \]

\[n - 1 = 15\ \ \]

\[n = 16\]

\[Ответ:В).\]

\[\mathbf{№11.}\]

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

\[d = a_{2} - a_{1} = 38 - 41 = - 3\]

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

\[41 - 3 \cdot (n - 1) \geq 0\ \]

\[41 - 3n + 3 \geq 0\]

\[44 \geq 3n \Longrightarrow \ \ n = \frac{44}{3} = 14\frac{2}{3}\]

\[\mathbf{Ответ:Б).}\]

\[\mathbf{№12.}\]

\[\left\{ \begin{matrix} a_{1} + a_{5} = 28 \\ a_{2} + a_{3} = 24 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} a_{1} + a_{1} + 4d = 28\ \ \ \ \ \ \ \\ a_{1} + d + a_{1} + 2d = 24 \\ \end{matrix} \right.\ \text{\ \ }\]

\[\ \left\{ \begin{matrix} 2a_{1} + 4d = 28 \\ 2a_{1} + 3d = 24 \\ \end{matrix} \right.\ ( - )\]

\[d = 4\]

\[\mathbf{Ответ:А).}\]

\[\mathbf{№13.}\]

\[\left\{ \begin{matrix} b_{2} = 24 \\ b_{5} = - 3 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} b_{1}q = 24\ \ \ \\ b_{1}q^{4} = - 3 \\ \end{matrix} \right.\ \ \ \ \ |\ :\]

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

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

\[q = - \frac{1}{2} \Longrightarrow \ \ b_{1} =\]

\[= 24\ :\left( - \frac{1}{2} \right) = - 48\]

\[S = \frac{- 48}{1 + \frac{1}{2}} = \frac{- 48}{\frac{3}{2}} =\]

\[= \frac{- 48 \cdot 2}{3} = - 32\]

\[Ответ:Г).\]

\[\mathbf{№14.}\]

\[0,(27) = 0,27 + 0,0027 + \ldots\]

\[q = \frac{0,0027}{0,27} = 0,01\]

\[0,(27) = \frac{0,27}{1 - 0,01} = \frac{0,27}{0,99} =\]

\[= \frac{27}{99} = \frac{3}{11}\]

\[\mathbf{Ответ:А).}\]

\[\mathbf{№15.}\]

\[a_{1} = 9,\ \ a_{2} = 18,\ \ \]

\[a_{n} = 117,\ \ d = 9\]

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

\[117 = 9 + 9 \cdot (n - 1)\ \]

\[117 = 9 + 9n - 9\]

\[9n = 117 \Longrightarrow \ \ n = 13\]

\[S = \frac{9 + 117}{2} \cdot 13 = \frac{126}{2} \cdot 13 =\]

\[= 63 \cdot 13 = 819\]

\[Ответ:\ В).\]

\[\mathbf{№16.}\]

\[a_{1} + a_{4} + a_{10} = 18\]

\[a_{1}\mathbf{+}a_{1} + a_{1} + 3d +\]

\[+ a_{1} + 9d = 18\]

\[3a_{1} + 12d = 18\ \ \ \ \ |\ \ :3\]

\[a_{1} + 4d = 6\]

\[S_{9} = \frac{2a_{1} + 8d}{2} \cdot 9 =\]

\[= \frac{2 \cdot \left( a_{1} + 4d \right)}{2} \cdot 9 =\]

\[= \frac{2 \cdot 6}{2} \cdot 9 = 6 \cdot 9 = 54\]

\[Ответ:\ \ Б).\]

\[\mathbf{№17.}\]

\[7x - 8,\ \ 2x + 1,\ \ x + 6\]

\[2 \cdot (2x + 1) =\]

\[= (7x - 8) + (x + 6)\]

\[4x + 2 = 7x + x - 8 + 6\]

\[- 4x = - 4\]

\[x = 1\]

\[Ответ:А).\]

\[\mathbf{№18.}\]

\[x + 1,\ \ 3x - 1,\ \ 2x + 10\]

\[(3x - 1)^{2} = (x + 1)(2x + 10)\]

\[9x^{2} - 6x + 1 =\]

\[= 2x^{2} + 10x + 2x + 10\]

\[7x^{2} - 18x - 9 = 0\]

\[D = 324 + 252 = 576\]

\[x = \frac{18 + 24}{14} = 3\]

\[x = \frac{18 - 24}{14} =\]

\[= - \frac{3}{7}\mathbf{\ (}\mathbf{не\ удовлетворяет).}\]

\[\mathbf{Ответ:В).}\]