Решебник по алгебре 9 класс Макарычев Задание 300, 299

 

\[\boxed{\text{300\ (299).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

\[Пусть\ \ t = x - \frac{1}{x};\ \ t^{2} =\]

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

\[t^{2} + 2 - \frac{1}{2}t = \frac{7}{2}\ \ \ \ \ \ \ \ \ \ \ \ \ | \cdot 2\]

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

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

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

\[t_{1} = \frac{1 + 5}{4} = \frac{3}{2};\ \ \ \ \ \ \]

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

\[1)\ при\ t_{1} = \frac{3}{2}:\]

\[x - \frac{1}{x} = \frac{3}{2}\ \ \ \ \ \ \ \ \ \ \ | \cdot 2x\]

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

\[D = 9 + 16 = 25\]

\[x_{1} = \frac{3 + 5}{4} = 2;\ \ \ \ \ \]

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

\[2)\ при\ t_{2} = - 1:\]

\[x - \frac{1}{x} = - 1\ \ \ \ \ \ \ \ \ \ | \cdot x\]

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

\[D = 1 + 4 = 5\]

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

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

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

\[Пусть\ t = x + \frac{1}{x};\ \ \ t^{2} =\]

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

\[t^{2} - 2 - \frac{1}{3}t = 8\ \ \ \ \ \ \ \ \ | \cdot 3\ \ \]

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

\[D = 1 + 360 = 361\]

\[t_{1} = \frac{1 - 19}{6} = - 3;\ t_{2} =\]

\[= \frac{1 + 19}{6} = \frac{10}{3}.\]

\[1)\ при\ t_{1} = - 3:\]

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

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

\[D = 9 - 4 = 5\]

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

\[2)\ при\ t_{2} = \frac{10}{3}:\]

\[x + \frac{1}{x} = \frac{10}{3}\ \ \ \ \ \ \ \ \ \ \ \ \ | \cdot 3x\ \]

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

\[D_{1} = 25 - 9 = 16\]

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

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