ГДЗ по алгебре 9 класс Макарычев ФГОС Задание 244

\[\boxed{\text{245.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ x² + \frac{1}{x²} - \frac{1}{2} \cdot \left( x - \frac{1}{x} \right) = 3\frac{1}{2}\]

\[Пусть\ \ t = x - \frac{1}{x},\ \ \]

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

\[\Longrightarrow t^{2} + 2 - \frac{1}{2}t = \frac{7}{2};\]

\[2t^{2} + 4 - t - 7 = 0\]

\[2t^{2} - t - 3 = 0\]

\[D = 1 + 4 \cdot 2 \cdot 3 = 25\]

\[t_{1,2} = \frac{1 \pm 5}{4} = \frac{3}{2};\ - 1;\]

\[1)\ при\ t_{1} = \frac{3}{2} \Longrightarrow x - \frac{1}{x} = \frac{3}{2},\ \]

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

\[2x^{2} - 3x - 2 = 0\]

\[D = 9 + 16 = 25\]

\[x_{1,2} = \frac{3 \pm 5}{4} = 2;\ - 0,5.\]

\[2)\ при\ t_{2} = - 1 \Longrightarrow x - \frac{1}{x} = - 1,\]

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

\[D = 1 + 4 = 5\]

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

\[Ответ:\ \ 2;\ - 0,5;\ \frac{- 1 \pm \sqrt{5}}{2}.\]

\[\textbf{б)}\ x² + \frac{1}{x^{2}} - \frac{1}{3} \cdot \left( x + \frac{1}{x} \right) = 8\]

\[Пусть\ t = x + \frac{1}{x},\]

\[\text{\ \ }t^{2} = \left( x + \frac{1}{x} \right)^{2} =\]

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

\[\Longrightarrow t^{2} - 2 - \frac{1}{3}t = 8,\]

\[\ \ 3t^{2} - t - 30 = 0,\]

\[t_{1,2} = - 3;\ \frac{10}{3}.\]

\[1)\ при\ t_{1} = - 3 \Longrightarrow x + \frac{1}{x} = - 3,\]

\[x² + 3x + 1 = 0\]

\[D = 9 - 4 = 5\]

\[x_{1.2} = \frac{- 3 \pm \sqrt{5}}{2};\ \]

\[2)\ при\ t_{2} = \frac{10}{3} \Longrightarrow x + \frac{1}{x} = \frac{10}{3},\ \]

\[3x^{2} - 10x + 3 = 0\]

\[x_{1} = 3,\ \ x_{2} = \frac{1}{3}.\]

\[Ответ:3;\frac{1}{3};\ \frac{- 3 \pm \sqrt{5}}{2}\text{.\ }\]