Решебник по алгебре 8 класс Макарычев ФГОС Задание 1278

\[\boxed{\text{1278.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[x^{2} - 2x - \frac{2}{x} + \frac{1}{x^{2}} - 13 = 0\]

\[\left( x^{2} + \frac{1}{x^{2}} \right) - \left( 2x - \frac{2}{x} \right) - 13 = 0\]

\[\left( x^{2} + \frac{1}{x^{2}} \right) - 2 \cdot\]

\[\cdot \left( x - \frac{1}{x} \right) - 13 = 0\]

\[\left( x^{2} + 2x \cdot \frac{1}{x} + \frac{1}{x^{2}} \right) - 2 -\]

\[- 2 \cdot \left( x + \frac{1}{x} \right) - 13 = 0\]

\[\left( x + \frac{1}{x} \right)^{2} - 2 \cdot \left( x + \frac{1}{x} \right) - 15 =\]

\[= 0,\ \ \left\lbrack t = x + \frac{1}{x} \right\rbrack\]

\[по\ т.\ Виета:\ \left\{ \begin{matrix} t_{1} + t_{2} = 2 \\ t_{1} \cdot t_{2} = - 15 \\ \end{matrix} \right.\ \]

\[t_{1} = 5\]

\[x + \frac{1}{x} = 5\]

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

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

\[D = 25 - 4 = 21\]

\[x_{1,2} = \frac{5 \pm \sqrt{21}}{2}\ \]

\[t_{2} = - 3\]

\[x + \frac{1}{x} = - 3\]

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

\[x^{2} + 3x + 1 = 0\]

\[D = 9 - 4 = 5\]

\[x_{3,4} = \frac{- 3 \pm \sqrt{5}}{2}\]

\[Ответ:x_{1,2} = \frac{5 \pm \sqrt{21}}{2};\ \]

\[\ x_{3,4} = \frac{- 3 \pm \sqrt{5}}{2}.\]