Решебник по алгебре 8 класс Макарычев ФГОС Задание 1264

\[\boxed{\text{1264.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[nx^{2} - 5x + 1 = 0\]

\[x_{1}^{- 2} + x_{2}^{- 2} = 13\]

\[x_{1}^{- 2} + x_{2}^{- 2} = \frac{1}{x_{1}^{2}} + \frac{1}{x_{2}^{2}} =\]

\[= \frac{x_{1}^{2} + x_{2}^{2}}{x_{1}^{2} \cdot x_{2}^{2}} =\]

\[= \frac{x_{1}^{2} + x_{2}^{2} + 2x_{1}x_{2} - 2x_{1}x_{2}}{\left( x_{1}x_{2} \right)^{2}} =\]

\[= \frac{\left( x_{1} + x_{2} \right)^{2} - 2x_{1}x_{2}}{(x_{1}x_{2})²} = 13\ (1)\]

\[nx² - 5x + 1 = 0\ \ |\ :n\]

\[x^{2} - \frac{5}{n}x + \frac{1}{n} = 0,\]

\[\text{\ \ }по\ т.\ Виета:\]

\[\left\{ \begin{matrix} x_{1} + x_{2} = \frac{5}{n} \\ x_{1} \cdot x_{2} = \frac{1}{n} \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow подставим\ в\ (1)\]

\[\frac{\left( x_{1} + x_{2} \right)^{2} - 2x_{1}x_{2}}{\left( x_{1}x_{2} \right)^{2}} = 13\]

\[\frac{\left( \frac{5}{n} \right)^{2} - 2 \cdot \frac{1}{n}}{\left( \frac{1}{n} \right)^{2}} = 13\]

\[\frac{\frac{25}{n^{2}} - \frac{2}{n}}{\frac{1}{n^{2}}} = 13\]

\[\frac{25 - 2n}{n^{2}} \cdot \frac{n^{2}}{1} = 13\]

\[25 - 2n = 13\]

\[- 2n = 13 - 25\]

\[- 2n = - 12\]

\[n = 6\]

\[Ответ:n = 6.\]