ГДЗ по алгебре 9 класс Макарычев ФГОС Задание 323

\[\boxed{\text{324.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ x^{4} - 9x^{2} + 18 = 0\]

\[Пусть\ x^{2} = t,\ t \geq 0\]

\[t^{2} - 9t + 18 = 0\]

\[D = 81 - 4 \cdot 18 = 9\]

\[t_{1,2} = \frac{9 \pm 3}{2} = 3;6;\ \]

\[1)\ x² = 3,\ \ x_{1,2} = \pm \sqrt{3};\]

\[2)\ x² = 6,\ \ x_{3,4} = \pm \sqrt{6}.\]

\[Сумма\ корней:\]

\[\sqrt{3} - \sqrt{3} + \sqrt{6} - \sqrt{6} = 0.\]

\[Ответ:0.\]

\[\textbf{б)}\ x^{4} + 3x² - 10 = 0\]

\[Пусть\ x^{2} = t,\ t \geq 0\]

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

\[D = 9 + 4 \cdot 10 = 49\]

\[t_{1,2} = \frac{- 3 \pm 7}{2} = - 5;2;\ \ так\ как\ \]

\[t \geq 0,\ то\ t = 2;\]

\[x^{2} = 2,\ \ \]

\[x_{1,2} = \pm \sqrt{2} \Longrightarrow\]

\[\Longrightarrow x_{1} + x_{2} = \sqrt{2} - \sqrt{2} = 0.\]

\[Ответ:0.\]

\[\textbf{в)}\ 4x^{4} - 12x^{2} + 1 = 0\]

\[Пусть\ t = x^{2},\ t \geq 0:\]

\[4t^{2} - 12t + 1 = 0\]

\[D = 36 - 4 = 32\]

\[t_{1,2} = \frac{6 \pm \sqrt{32}}{4} = \frac{6 \pm 4\sqrt{2}}{4} =\]

\[= \frac{3 \pm \sqrt{2}}{2};\]

\[1)\ x² = \frac{3 + \sqrt{2}}{2},\ \ \]

\[x_{1,2} = \pm \sqrt{\frac{3 + \sqrt{2}}{2}};\]

\[2)\ x² = \frac{3 - \sqrt{2}}{2},\ \ \]

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

\[Ответ:0.\]

\[\textbf{г)}\ 12y^{4} - y^{2} - 1 = 0\]

\[Пусть\ \ t = y^{2},\ t \geq 0:\]

\[12t^{2} - t - 1 = 0\]

\[D = 1 + 4 \cdot 12 = 49\]

\[t_{1,2} = \frac{1 \pm 7}{24};\]

\[так\ как\ t \geq 0,\ то\ t = \frac{1}{3};\]

\[y^{2} = \frac{1}{3},\ \ y_{1,2} = \pm \sqrt{\frac{1}{3}};\]

\[y_{1} + y_{2} = \sqrt{\frac{1}{3}} - \sqrt{\frac{1}{3}} = 0.\]

\[Ответ:0.\]