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МатематикаРешебник по алгебре 9 класс Мерзляк Задание 136
Задание 136
\[\boxed{\text{136\ (136).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[1)\ 7(x + 2) - 3(x - 8) < 10\]
\[7x + 14 - 2x + 24 < 10\]
\[4x < - 28\]
\[x < - 7\ \ \]
\[x \in ( - \infty;\ - 7)\]
\[x_{\max} = - 8.\]
\[2)\ (x - 4)(x + 4) -\]
\[- 5x > (x - 1)^{2} - 17\]
\[x^{2} - 16 - 5x > x^{2} - 2x + 1 - 17\]
\[x^{2} - x^{2} - 5x + 2x > 1 - 17 + 16\]
\[- 3x > 0\]
\[x < 0\]
\[x \in ( - \infty;0)\]
\[x_{\max} = - 1.\]
\[\boxed{\mathbf{136.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[1)\text{\ x}^{4} - 29x^{2} + 100 = 0\]
\(\left\{ \begin{matrix} x^{2} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ t² - 29t + 100 = 0 \\ \end{matrix} \right.\ \)
\(t_{1} + t_{2} = 29,\ \ t_{1} \cdot t_{2} = 100,\)
\[\text{\ \ }t_{1} = 25,\ \ t_{2} = 4\]
\[Ответ:\ x = \pm 5;x = \pm 2.\ \]
\[2)\ \text{\ x}^{4} - 9x^{2} + 20 = 0\]
\(\left\{ \begin{matrix} x^{2} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ t² - 9t + 20 = 0 \\ \end{matrix} \right.\ \)
\(t_{1} + t_{2} = 9,\ \ t_{1} \cdot t_{2} = 20,\ \ \)
\[t_{1} = 5,\ \ t_{2} = 4\]
\[Ответ:\ x = \pm \sqrt{5};x = \ \pm 2.\]
\[3)\text{\ x}^{4} - 2x^{2} - 24 = 0\]
\(\left\{ \begin{matrix} x^{2} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ t² - 2t - 24 = 0 \\ \end{matrix} \right.\ \)
\[t_{1} + t_{2} = 2,\ \ t_{1} \cdot t_{2} = - 24,\ \]
\[\ t_{1} = 6,\ \ t_{2} = - 4\]
\[Ответ:\ x = \pm \sqrt{6}.\]
\[4)\ \text{\ x}^{4} + 3x^{2} - 70 = 0\]
\(\left\{ \begin{matrix} x^{2} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ t² + 3t - 70 = 0 \\ \end{matrix} \right.\ \)
\(t_{1} + t_{2} = - 3,\ \ t_{1} \cdot t_{2} = - 70,\ \ \)
\[t_{1} = - 10,\ \ t_{2} = 7\]
\[Ответ:\ x = \pm \sqrt{7}.\]
\[5)\ {\ 9x}^{4} - 10x^{2} + 1 = 0\]
\(\left\{ \begin{matrix} x^{2} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 9t² - 10t + 1 = 0 \\ \end{matrix} \right.\ \)
\[D = 100 - 36 = 64,\ \ \]
\[t_{1} = \frac{10 + 8}{18} = 1,\ \ \]
\[t_{2} = \frac{10 - 8}{18} = \frac{1}{9}\]
\[Ответ:x = \pm 1;x = \pm \frac{1}{3}.\]
\[6){\ 2x}^{4} - 5x^{2} + 2 = 0\]
\(\left\{ \begin{matrix} x^{2} = t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2t² - 5t + 2 = 0 \\ \end{matrix} \right.\ \)
\[D = 25 - 16 = 9,\]
\[\text{\ \ }t_{1} = \frac{5 - 3}{4} = \frac{1}{2},\ \ \]
\[t_{2} = \frac{5 + 3}{4} = 2\]
\[Ответ:x = \pm \frac{1}{\sqrt{2}};\ x = \pm \sqrt{2}\text{.\ }\]
