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МатематикаРешебник по алгебре 9 класс Макарычев Задание 565
Задание 565
\[\boxed{\text{565\ (565).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ x_{n} = 2n - 1\]
\[x_{1} = 2 - 1 = 1\]
\[x_{2} = 4 - 1 = 3\]
\[x_{3} = 6 - 1 = 5\]
\[x_{4} = 8 - 1 = 7\]
\[x_{5} = 10 - 1 = 9\]
\[x_{6} = 12 - 1 = 11;\]
\[\textbf{б)}\ x_{n} = n^{2} + 1\]
\[\ x_{1} = 1 + 1 = 2\]
\[x_{2} = 4 + 1 = 5\]
\[x_{3} = 9 + 1 = 10\]
\[x_{4} = 16 + 1 = 17\]
\[x_{5} = 25 + 1 = 26\]
\[x_{6} = 36 + 1 = 37\]
\[\textbf{в)}\ x_{n} = \frac{n}{n + 1}\]
\[x_{1} = \frac{1}{1 + 1} = \frac{1}{2}\]
\[x_{2} = \frac{2}{2 + 1} = \frac{2}{3}\]
\[x_{3} = \frac{3}{3 + 1} = \frac{3}{4}\]
\[x_{4} = \frac{4}{4 + 1} = \frac{4}{5}\]
\[x_{5} = \frac{5}{5 + 1} = \frac{5}{6}\]
\[x_{6} = \frac{6}{6 + 1} = \frac{6}{7}\ \]
\[\textbf{г)}\ x^{n} = ( - 1)^{n + 1} \cdot 2\ \]
\[x_{1} = ( - 1)^{2} \cdot 2 = 2\]
\[x_{2} = - 2\]
\[x_{3} = 2\]
\[x_{4} = - 2\]
\[x_{5} = 2\]
\[x_{6} = - 2;\]
\[\textbf{д)}\ x_{n} = 2^{n - 3}\ \]
\[x_{1} = 2^{- 2} = \frac{1}{4}\]
\[x_{2} = 2^{- 1} = \frac{1}{2}\]
\[x_{3} = 2^{0} = 1\]
\[x_{4} = 2^{1} = 2\]
\[x_{5} = 2^{2} = 4\]
\[x_{6} = 2³ = 8;\]
\[\textbf{е)}\ x^{n} = 0,5 \cdot 4^{n}\ \]
\[x_{1} = 0,5 \cdot 4 = 2\]
\[x_{2} = 0,5 \cdot 4^{2} = 8\]
\[x_{3} = 0,5 \cdot 4^{3} = 32\]
\[x_{4} = 0,5 \cdot 4^{4} = 128\]
\[x_{5} = 0,5 \cdot 4^{5} = 512\]
\[x_{6} = 0,5 \cdot 4^{6} = 2048.\]
\[\boxed{\text{565.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\left\{ \begin{matrix} 3x + y = 2\ \ \ \ \ \ \ \ \\ x^{2} - y^{2} = - 12 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = 2 - 3x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - (2 - 3x)^{2} = - 12 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = 2 - 3x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 4 + 12x - 9x^{2} = - 12 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = 2 - 3x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 8x^{2} - 12x - 8 = 0 \\ \end{matrix} \right.\ \]
\[2x^{2} - 3x - 2 = 0\]
\[D = 9 + 4 \cdot 2 \cdot 2 = 25\]
\[x_{1,2} = \frac{3 \pm 5}{4};\]
\[\left\{ \begin{matrix} x_{1} = 2\ \ \ \\ y_{1} = - 4 \\ \end{matrix} \right.\ \ \ \ \ \ или\ \ \ \left\{ \begin{matrix} x_{2} = - \frac{1}{2} \\ y_{2} = 3,5. \\ \end{matrix} \right.\ \]
\[Ответ:(2;\ - 4);\ \ ( - 0,5;3,5).\]
