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МатематикаГДЗ по алгебре 9 класс Макарычев Задание 847
Задание 847
\[\boxed{\text{847\ (847).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[Пусть\ в\ группе\ было\ \text{x\ }\]
\[туристов:\]
\[C_{x}^{2} = \frac{x!}{(x - 2)!} =\]
\[= \frac{(x - 2)!(x - 1) \cdot x}{(x - 2)!} =\]
\[= (x - 1) \cdot x.\]
\[Если\ бы\ было\ на\ одного\ \]
\[человека\ больше,\ то:\]
\[C_{x + 1}^{2} = \frac{(x + 1)!}{\left( (x + 1) - 2 \right)!} =\]
\[= \frac{(x + 1)!}{(x - 1)!} =\]
\[= \frac{(x - 1)!\left( x \cdot (x + 1) \right)!}{(x - 1)!} =\]
\[= x(x + 1).\]
\[\frac{A_{x + 1}^{2}}{A_{x}^{2}} = 1,25\]
\[\frac{(x + 1)x}{x(x - 1)} = 1\frac{1}{4}\]
\[\frac{x + 1}{x - 1} = \frac{5}{4}\]
\[4 \cdot (x + 1) = 5 \cdot (x - 1)\]
\[4x + 4 = 5x - 5\]
\[- x = - 9\]
\[x = 9\]
\[Ответ:9\ туристов.\]
\[\boxed{\text{847.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[x^{2} - 3ax + a^{2} = 0,\]
\[x_{1}^{2} + x_{2}^{2} = 1,75;\]
\[x_{1} - ?\ \ \ \ x_{2} - ?\]
\[По\ теореме\ Виета:\]
\[x_{1} + x_{2} = 3a,\]
\[x_{1} \cdot x_{2} = a^{2},\]
\[\left( x_{1} \right)^{2} + \left( x_{2} \right)^{2} =\]
\[= \left( x_{1} + x_{2} \right)^{2} - 2x_{1}x_{2} = 1,75;\]
\[9a^{2} - 2a^{2} = 1,75;\]
\[7a^{2} = 1,75\]
\[a^{2} = 0,25 \Longrightarrow a = \pm 0,5.\]
\[1)\ \left\{ \begin{matrix} x_{1} + x_{2} = 1,5 \\ x_{1}x_{2} = 0,25\ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x_{2} = 1,5 - x_{1}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ 1,5x_{1} - x_{1}^{2} - 0,25 = 0 \\ \end{matrix} \right.\ \]
\[4x_{1}^{2} - 6x_{1} + 1 = 0\]
\[D = 36 - 16 = 20 = 2\sqrt{5},\]
\[x_{1,2} = \frac{6 \pm 2\sqrt{5}}{8} = \frac{3 \pm \sqrt{5}}{4}\text{.\ }\]
\[\left\{ \begin{matrix} x_{1} = \frac{3 + \sqrt{5}}{4} \\ x_{2} = \frac{3 - \sqrt{5}}{4} \\ \end{matrix} \right.\ \text{\ \ \ }или\]
\[\text{\ \ \ }\left\{ \begin{matrix} x_{1} = \frac{3 - \sqrt{5}}{4} \\ x_{2} = \frac{3 + \sqrt{5}}{4} \\ \end{matrix} \right.\ .\]
\[2)\ \ \left\{ \begin{matrix} x_{1} + x_{2} = - 1,5 \\ x_{1}x_{2} = 0,25\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x_{2} = - 1,5 - x_{1}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ - 1,5x_{1} - x_{1}^{2} - 0,25 = 0 \\ \end{matrix} \right.\ \]
\[4x_{1}^{2} + 6x_{1} + 1 = 0\]
\[D = 36 - 16 = 20 = 2\sqrt{5},\]
\[x_{1,2} = \frac{- 6 \pm 2\sqrt{5}}{8} = \frac{- 3 \pm \sqrt{5}}{4}\text{.\ }\]
\[\left\{ \begin{matrix} x_{1} = \frac{- 3 + \sqrt{5}}{4} \\ x_{2} = \frac{- 3 - \sqrt{5}}{4} \\ \end{matrix} \right.\ \text{\ \ \ }или\]
\[\text{\ \ \ }\left\{ \begin{matrix} x_{1} = \frac{- 3 - \sqrt{5}}{4} \\ x_{2} = \frac{- 3 + \sqrt{5}}{4} \\ \end{matrix} \right.\ .\]
\[Ответ:x_{1} = \frac{- 3 - \sqrt{5}}{4},\ \frac{- 3 + \sqrt{5}}{4},\ \ \]
\[x_{2} = \frac{- 3 + \sqrt{5}}{4},\ \frac{- 3 - \sqrt{5}}{4}.\]
\[\boxed{\text{848.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[x^{2} - 3,75x + a^{3} = 0,\ \ a - ?\]
\[По\ теореме\ Виета:\ \]
\[x_{1} + x_{2} = 3,75,\]
\[x_{1} \cdot x_{2} = a³.\]
\[x_{1}x_{2} = x_{2}^{2}x_{2} = x_{2}^{3} = a^{3},\]
\[x_{2} = a,\ \ x_{1} = a^{2}.\]
\[a^{2} + a = 3,75\]
\[a^{2} + a - 3,75 = 0\]
\[D = 1 + 15 = 16\]
\[a_{1} = \frac{- 1 + 4}{2} = 1,5,\ \ \]
\[a_{2} = \frac{- 1 - 4}{2} = - 2,5.\]
\[Ответ:a_{1} = 1,5;\ \ \ a_{2} = - 2,5.\]
