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МатематикаГДЗ по алгебре 9 класс Макарычев ФГОС Задание 222
Задание 222
\[\boxed{\text{223.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ x^{4} - 5x^{2} - 36 = 0\]
\[Пусть\ \ x^{2} = t,\ x^{4} = t^{2},\ \ t \geq 0.\]
\[t^{2} - 5t - 36 = 0\]
\[D = 25 + 4 \cdot 36 = 169\]
\[t_{1,2} = \frac{5 \pm 13}{2}\text{\ \ }\]
\[t_{1} = 9\ \ \ \ ;\ \ \ \ \ t_{2} = - 4 \Longrightarrow\]
\[\Longrightarrow не\ подходит.\]
\[x^{2} = 9\ \]
\[x = \pm 3.\]
\[Ответ:x = \pm 3.\]
\[\textbf{б)}\ y^{4} - 6y^{2} + 8 = 0\]
\[Пусть\ \ y^{2} = t,\ \ y^{4} = t^{2},\]
\[\ \ t \geq 0.\]
\[t^{2} - 6t + 8 = 0\]
\[D = 9 - 8 = 1\]
\[t_{1,2} = 3 \pm 1\]
\[t_{1} = 4;\ \ \ \ \ t_{2} = 2;\]
\[\left\{ \begin{matrix} y² = 4 \\ y² = 2 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y_{1,2} = \pm 2 \\ y_{3,4} = \pm \sqrt{2} \\ \end{matrix} \right.\ .\]
\[Ответ:y = \pm 2;\ \ y = \pm \sqrt{2}.\]
\[\textbf{в)}\ t^{4} + 10t² + 25 = 0\]
\[Пусть\ t^{2} = a,\ \ t^{4} = a^{2},\]
\[\ \ a \geq 0.\]
\[a^{2} + 10a + 25 = 0\]
\[(a + 5)^{2} = 0\]
\[a = - 5 \Longrightarrow не\ подходит.\]
\[Ответ:корней\ исходное\ \]
\[уравнение\ не\ имеет.\]
\[\textbf{г)}\ 4x^{4} - 5x^{2} + 1 = 0\]
\[Пусть\ x^{2} = a,\ \ x^{4} = a^{2},\]
\[\ \ a \geq 0.\]
\[4a^{2} - 5a + 1 = 0\]
\[D = 25 - 16 = 9\]
\[a_{1,2} = \frac{5 \pm 3}{8} = 1;\frac{1}{4}.\]
\[\left\{ \begin{matrix} x² = 1 \\ x² = \frac{1}{4} \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x_{1,2} = \pm 1 \\ x_{3,4} = \pm \frac{1}{2} \\ \end{matrix} \right.\ \]
\[Ответ:x = \pm 1;\ \ x = \pm \frac{1}{2}.\]
\[\textbf{д)}\ 9x^{4} - 9x^{2} + 2 = 0\]
\[Пусть\ \ x^{2} = t,\ \ t \geq 0.\]
\[9t^{2} - 9t + 2 = 0\]
\[D = 9\]
\[t_{1,2} = \frac{9 \pm 3}{18} = \frac{1}{3};\ \frac{2}{3}.\]
\[x^{2} = \frac{1}{3} \Longrightarrow x = \pm \sqrt{\frac{1}{3}}.\]
\[x² = \frac{2}{3} \Longrightarrow x = \pm \sqrt{\frac{2}{3}}.\]
\[Ответ:\ x = \pm \sqrt{\frac{2}{3}};\ \ x = \pm \sqrt{\frac{1}{3}}.\]
\[\textbf{е)}\ 16y^{4} - 8y^{2} + 1 = 0\]
\[Пусть\ \ y^{2} = t,\ \ t \geq 0.\]
\[16t^{2} - 8t + 1 = 0\]
\[(4t - 1)^{2} = 0\]
\[t = \frac{1}{4} \Longrightarrow y^{2} = \frac{1}{4} \Longrightarrow\]
\[y = \pm \frac{1}{2}.\]
\[Ответ:y = \pm \frac{1}{2}.\]
