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МатематикаГДЗ по алгебре 8 класс Мерзляк Задание 775
Задание 775
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\[1)\ x^{4} - 5x^{2} + 4 = 0\]
\(\left\{ \begin{matrix} x^{2} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ t² - 5t + 4 = 0 \\ \end{matrix} \right.\ \)
\(t_{1} + t_{2} = 5,\ \ t_{1} \cdot t_{2} = 4,\ \ \)
\[t_{1} = 4,\ \ t_{2} = 1\]
\[\left\{ \begin{matrix} (t - 4)(t - 1) = 0 \\ x^{2} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[Ответ:\ x = \pm 2;\ x = \pm 1.\]
\[2)\ x^{4} - 5x^{2} + 6 = 0\]
\(\left\{ \begin{matrix} x^{2} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ t² - 5t + 6 = 0 \\ \end{matrix} \right.\ \)
\(t_{1} + t_{2} = 5,\ \ t_{1} \cdot t_{2} = 6,\ \ \)
\[t_{1} = 3,\ \ t_{2} = 2\]
\[Ответ:x = \pm \sqrt{3};\ x = \pm \sqrt{2}.\ \]
\[3)\ x^{4} - 8x^{2} - 9 = 0\]
\(\left\{ \begin{matrix} x^{2} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ t² - 8t - 9 = 0 \\ \end{matrix} \right.\ \)
\[t_{1} + t_{2} = 8,\ \ t_{1} \cdot t_{2} = - 9,\ \ \]
\[t_{1} = 9,\ \ t_{2} = - 1\]
\[Ответ:x = 3;\ x = - 3.\ \]
\[4)\ x^{4} + 14x^{2} - 32 = 0\]
\(\left\{ \begin{matrix} x^{2} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ t² + 14t - 32 = 0 \\ \end{matrix} \right.\ \)
\(t_{1} + t_{2} = - 14,\ \ t_{1} \cdot t_{2} = - 32,\ \ \)
\[t_{1} = - 16,\ \ t_{2} = 2\]
\[\left\{ \begin{matrix} (t + 16)(t - 2) = 0 \\ x^{2} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[Ответ:x = \pm \sqrt{2}\text{.\ }\]
\[5)\ 4x^{4} - 9x^{2} + 2 = 0\]
\(\left\{ \begin{matrix} x^{2} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 4t² - 9t + 2 = 0 \\ \end{matrix} \right.\ \)
\[D = 81 - 32 = 49,\ \ \]
\[t_{1} = \frac{9 + 7}{8} = 2,\ \ \]
\[t_{2} = \frac{9 - 7}{8} = \frac{1}{4}\]
\[\left\{ \begin{matrix} 4 \cdot (t - 2)\left( t - \frac{1}{4} \right) = 0 \\ x^{2} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[Ответ:x = \pm \frac{1}{2};x = \pm \sqrt{2}\text{.\ }\]
\[6)\ {3x}^{4} + 8x^{2} - 3 = 0\]
\[\left\{ \begin{matrix} x^{2} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3t² + 8t - 3 = 0 \\ \end{matrix} \right.\ \]
\[D = 64 + 36 = 100,\ \ \]
\[t_{1} = \frac{- 8 + 10}{6} = \frac{1}{3},\ \ \]
\[t_{2} = \frac{- 8 - 10}{6} = - 3\]
\[\left\{ \begin{matrix} 3 \cdot \left( t - \frac{1}{3} \right)(t + 3) = 0 \\ x^{2} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[Ответ:x = \pm \frac{1}{\sqrt{3}}\text{.\ }\]
