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МатематикаРешебник по алгебре 9 класс Мерзляк Проверь себя №6
Проверь себя №6
\[\boxed{\mathbf{Задание}\mathbf{\ }\mathbf{№}\mathbf{6.}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[\mathbf{№1.}\]
\[\mathbf{Ответ:Б).}\]
\[\mathbf{№2.}\]
\[\mathbf{Ответ:В).}\]
\[\mathbf{№3.}\]
\[a_{6} = a_{1} + 5d = 12 +\]
\[+ 5 \cdot 0,4 = 14\]
\[\mathbf{Ответ:Б).}\]
\[\mathbf{№4.}\]
\[a_{2}\mathbf{=}a_{1} + d\]
\[d = a_{2} - a_{1} = 5 + 7 = 12\]
\[\mathbf{Ответ:Г).}\]
\[\mathbf{№5.}\]
\[S_{10} = \frac{2a_{1} + 9d}{2} \cdot 10 =\]
\[= \left( 2a_{1} + 9d \right) \cdot 5 =\]
\[= ( - 32 + 27) \cdot 5 = - 25\]
\[\mathbf{Ответ:Г).}\]
\[\mathbf{№6.}\]
\[b_{n} = b_{1}q^{3} = - \frac{1}{8} \cdot ( - 2)^{3} =\]
\[= - \frac{1}{8} \cdot ( - 8) = 1\]
\[\mathbf{Ответ:В).}\]
\[\mathbf{№7.}\]
\[b_{2} = b_{1}q,\ \ q = \frac{b_{2}}{b_{1}} = \frac{9}{36} = \frac{1}{4}\]
\[\mathbf{Ответ:А).}\]
\[\mathbf{№8.}\]
\[S_{n} = \frac{2 \cdot \left( 3^{4} - 1 \right)}{3 - 1} =\]
\[= \frac{2 \cdot (81 - 1)}{2} = 80\]
\[Ответ:Б).\]
\[\mathbf{№9.}\]
\[a_{n} = - 4n + 13\ \]
\[a_{1} = - 4 + 13 = 9\]
\[a = - 4 \cdot 2 + 13 = 5\]
\[a_{15} = - 4 \cdot 15 + 13 = - 47\]
\[S_{15} = \frac{9 - 47}{2} \cdot 15 =\]
\[= - 19 \cdot 15 = - 285\]
\[Ответ:Б).\]
\[\mathbf{№10.}\]
\[a_{n} = a_{1} + d(n - 1)\text{\ \ }\]
\[6,2 = 0,2 + 0,4 \cdot (n - 1)\ \]
\[\ 0,4 \cdot (n - 1) = 6\ \ \]
\[n - 1 = 15\ \ \]
\[n = 16\]
\[Ответ:В).\]
\[\mathbf{№11.}\]
\[a_{2} = a_{1} + d\text{\ \ }\]
\[d = a_{2} - a_{1} = 38 - 41 = - 3\]
\[a_{n} = a_{1} + d(n - 1)\]
\[41 - 3 \cdot (n - 1) \geq 0\ \]
\[41 - 3n + 3 \geq 0\]
\[44 \geq 3n \Longrightarrow \ \ n = \frac{44}{3} = 14\frac{2}{3}\]
\[\mathbf{Ответ:Б).}\]
\[\mathbf{№12.}\]
\[\left\{ \begin{matrix} a_{1} + a_{5} = 28 \\ a_{2} + a_{3} = 24 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} a_{1} + a_{1} + 4d = 28\ \ \ \ \ \ \ \\ a_{1} + d + a_{1} + 2d = 24 \\ \end{matrix} \right.\ \text{\ \ }\]
\[\ \left\{ \begin{matrix} 2a_{1} + 4d = 28 \\ 2a_{1} + 3d = 24 \\ \end{matrix} \right.\ ( - )\]
\[d = 4\]
\[\mathbf{Ответ:А).}\]
\[\mathbf{№13.}\]
\[\left\{ \begin{matrix} b_{2} = 24 \\ b_{5} = - 3 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} b_{1}q = 24\ \ \ \\ b_{1}q^{4} = - 3 \\ \end{matrix} \right.\ \ \ \ \ |\ :\]
\[\frac{1}{q^{3}} = - 8\ \ \]
\[q^{3} = - \frac{1}{8}\text{\ \ }\]
\[q = - \frac{1}{2} \Longrightarrow \ \ b_{1} =\]
\[= 24\ :\left( - \frac{1}{2} \right) = - 48\]
\[S = \frac{- 48}{1 + \frac{1}{2}} = \frac{- 48}{\frac{3}{2}} =\]
\[= \frac{- 48 \cdot 2}{3} = - 32\]
\[Ответ:Г).\]
\[\mathbf{№14.}\]
\[0,(27) = 0,27 + 0,0027 + \ldots\]
\[q = \frac{0,0027}{0,27} = 0,01\]
\[0,(27) = \frac{0,27}{1 - 0,01} = \frac{0,27}{0,99} =\]
\[= \frac{27}{99} = \frac{3}{11}\]
\[\mathbf{Ответ:А).}\]
\[\mathbf{№15.}\]
\[a_{1} = 9,\ \ a_{2} = 18,\ \ \]
\[a_{n} = 117,\ \ d = 9\]
\[a_{n} = a_{1} + d(n - 1)\text{\ \ }\]
\[117 = 9 + 9 \cdot (n - 1)\ \]
\[117 = 9 + 9n - 9\]
\[9n = 117 \Longrightarrow \ \ n = 13\]
\[S = \frac{9 + 117}{2} \cdot 13 = \frac{126}{2} \cdot 13 =\]
\[= 63 \cdot 13 = 819\]
\[Ответ:\ В).\]
\[\mathbf{№16.}\]
\[a_{1} + a_{4} + a_{10} = 18\]
\[a_{1}\mathbf{+}a_{1} + a_{1} + 3d +\]
\[+ a_{1} + 9d = 18\]
\[3a_{1} + 12d = 18\ \ \ \ \ |\ \ :3\]
\[a_{1} + 4d = 6\]
\[S_{9} = \frac{2a_{1} + 8d}{2} \cdot 9 =\]
\[= \frac{2 \cdot \left( a_{1} + 4d \right)}{2} \cdot 9 =\]
\[= \frac{2 \cdot 6}{2} \cdot 9 = 6 \cdot 9 = 54\]
\[Ответ:\ \ Б).\]
\[\mathbf{№17.}\]
\[7x - 8,\ \ 2x + 1,\ \ x + 6\]
\[2 \cdot (2x + 1) =\]
\[= (7x - 8) + (x + 6)\]
\[4x + 2 = 7x + x - 8 + 6\]
\[- 4x = - 4\]
\[x = 1\]
\[Ответ:А).\]
\[\mathbf{№18.}\]
\[x + 1,\ \ 3x - 1,\ \ 2x + 10\]
\[(3x - 1)^{2} = (x + 1)(2x + 10)\]
\[9x^{2} - 6x + 1 =\]
\[= 2x^{2} + 10x + 2x + 10\]
\[7x^{2} - 18x - 9 = 0\]
\[D = 324 + 252 = 576\]
\[x = \frac{18 + 24}{14} = 3\]
\[x = \frac{18 - 24}{14} =\]
\[= - \frac{3}{7}\mathbf{\ (}\mathbf{не\ удовлетворяет).}\]
\[\mathbf{Ответ:В).}\]
