ГДЗ по алгебре 9 класс Макарычев Задание 358

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Год:2020-2021-2022
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Задание 358

\[\boxed{\text{358\ (358).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \left( x^{2} + 6x \right)^{2} -\]

\[- 5 \cdot \left( x^{2} + 6x \right) = 24\]

\[Пусть\ t = x^{2} + 6x:\ \ \]

\[t^{2} - 5t = 24\]

\[t^{2} - 5t - 24 = 0\]

\[D = 25 + 4 \cdot 24 = 121\]

\[t_{1,2} = \frac{5 \pm 11}{2} = - 3;8;\]

\[1)\ x^{2} + 6x = - 3\]

\[x^{2} + 6x + 3 = 0\]

\[D_{1} = 9 - 3 = 6\]

\[x_{1,2} = - 3 \pm \sqrt{6}.\]

\[2)\ x^{2} + 6x = 8\]

\[x^{2} + 6x - 8 = 0\]

\[D_{1} = 9 + 8 = 17\]

\[x_{3,4} = - 3 \pm \sqrt{17}.\]

\[Ответ: - 3 \pm \sqrt{17};\ - 3 \pm \sqrt{6}.\]

\[\textbf{б)}\ \left( x^{2} - 2x - 5 \right)^{2} -\]

\[- 2 \cdot \left( x^{2} - 2x - 5 \right) = 3\]

\[Пусть\ t = x^{2} - 2x - 5:\ \]

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

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

\[D_{1} = 1 + 3 = 4\]

\[t_{1,2} = 1 \pm 2 = - 1;3.\]

\[1)\ x^{2} - 2x - 5 = - 1\]

\[x^{2} - 2x - 4 = 0\]

\[D_{1} = 1 + 4 = 5\]

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

\[2)\ x^{2} - 2x - 5 = 3\]

\[x^{2} - 2x - 8 = 0\]

\[D_{1} = 1 + 8 = 9\]

\[x_{1,2} = 1 \pm 3 = 4;\ - 2.\]

\[Ответ: - 2;4;1 \pm \sqrt{5}.\]

\[\textbf{в)}\ \left( x^{2} + 3x - 25 \right)^{2} -\]

\[- 2 \cdot \left( x^{2} + 3x - 25 \right) = - 7\]

\[Пусть\ x^{2} + 3x - 25 = t:\]

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

\[D_{1} = 1 - 7 = - 6 < 0 \Longrightarrow\]

\[\Longrightarrow корней\ нет.\]

\[Ответ:\ корней\ нет.\]

\[\textbf{г)}\ (y + 2)^{4} - (y + 2)^{2} = 12\]

\[Пусть\ t = (y + 2)^{2};\ t \geq 0:\ \]

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

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

\[t_{1,2} = \frac{1 \pm 7}{2} = 4; - 3.\]

\[так\ как\ \ t \geq 0:\]

\[(y + 2)^{2} = 4\]

\[y + 2 = \pm 2;\]

\[1)\ y + 2 = 2\ \ \]

\[y_{1} = 0.\]

\[2)\ y + 2 = - 2\ \ \]

\[y_{2} = - 4.\]

\[Ответ:\ - 4;0.\]

\[\textbf{д)}\ \left( x^{2} + 2x \right)\left( x^{2} + 2x + 2 \right) = 3\]

\[Пусть\ t = x^{2} + 2x:\ \]

\[t(t + 2) = 3\]

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

\[D_{1} = 1 + 3 = 4\]

\[t_{1,2} = - 1 \pm 2 = - 3;1.\]

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

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

\[D_{1} = 1 - 3 < 0 \Longrightarrow корней\ нет;\]

\[2)\ x^{2} + 2x = 1\]

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

\[D_{1} = 1 + 1 = 2\]

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

\[Ответ: - 1 \pm \sqrt{2}.\]

\[\textbf{е)}\ \left( x^{2} - x - 16 \right)\left( x^{2} - x + 2 \right) = 88\]

\[Пусть\ t = x^{2} - x:\ \]

\[(t - 16)(t + 2) = 88\]

\[t^{2} - 16t + 2t - 32 - 88 = 0\]

\[t^{2} - 14t - 120 = 0\]

\[D_{1} = 49 + 120 = 169\]

\[t_{1,2} = 7 \pm 13 = 20;\ - 6.\]

\[1)\ x^{2} - x = 20\]

\[x^{2} - x - 20 = 0\]

\[D = 1 + 4 \cdot 20 = 81\]

\[x_{1,2} = \frac{1 \pm 9}{2} = 5; - 4.\]

\[2)\ x^{2} - x = - 6\]

\[x^{2} - x + 6 = 0\]

\[D = 1 - 4 \cdot 6 < 0 \Longrightarrow корней\ нет.\]

\[Ответ: - 4;5.\]

\[\textbf{ж)}\ \left( 2x^{2} + 7x - 8 \right) \cdot\]

\[\cdot \left( 2x^{2} + 7x - 3 \right) - 6 = 0\]

\[Пусть\ \ t = 2x^{2} + 7x:\]

\[(t - 8)(t - 3) - 6 = 0\]

\[t^{2} - 11t + 18 = 0\]

\[t_{1} + t_{2} = 11;\ \ \ \ t_{1} \cdot t_{2} = 18\]

\[t_{1} = 2;\ \ t_{2} = 9.\]

\[1)\ 2x^{2} + 7x = 2\]

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

\[D = 49 + 4 \cdot 2 \cdot 2 = 65\]

\[x_{1,2} = \frac{- 7 \pm \sqrt{65}}{4}.\]

\[2)\ 2x^{2} + 7x = 9\]

\[2x^{2} + 7x - 9 = 0\]

\[D = 49 + 4 \cdot 2 \cdot 9 = 121\]

\[x_{3,4} = \frac{- 7 \pm 11}{4} = 1;\ - 4,5.\]

\[Ответ:\ - 4,5;1;\ \frac{- 7 \pm \sqrt{65}}{4}.\]

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