Алгебра
Английский
Биология
География
Геометрия
История
Русский
Обществознание
Физика
Химия
Информатика
Литература
МатематикаРешебник по алгебре 9 класс Макарычев Задание 321
Задание 321
\[\boxed{\text{321\ (321).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ y = \sqrt{25 - x^{2}} + \sqrt{9x - x^{2} - 14}\]
\[ОДЗ:\]
\[\left\{ \begin{matrix} 25 - x^{2} \geq 0\ \ \ \ \ \ \ \ \ \ \\ 9x - x^{2} - 14 \geq 0 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} (x - 5)(x + 5) \leq 0 \\ x^{2} - 9x + 14 \leq 0\ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} (x - 5)(x + 5) \leq 0 \\ (x - 2)(x - 7) \leq 0 \\ \end{matrix} \right.\ \]
\[x^{2} - 9x + 14 = 0\]
\[x_{1} + x_{2} = 9;\ \ \ x_{1} \cdot x_{2} = 14\]
\[x_{1} = 2;\ \ \ \ \ \ \ \ \ \ \ x_{2} = 7\]
\[\ x \in \lbrack 2;5\rbrack.\]
\[Ответ:\ x = 2;3;4;5.\]
\[\textbf{б)}\ y = \sqrt{8x - x^{2} - 12} +\]
\[+ \sqrt{16 - x^{2}}\]
\[ОДЗ:\]
\[\left\{ \begin{matrix} 8x - x^{2} - 12 \geq 0 \\ 16 - x^{2} \geq 0\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x^{2} - 8x + 12 \leq 0 \\ x^{2} - 16 \leq 0\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} (x - 2)(x - 6) \leq 0 \\ (x - 4)(x + 4) \leq 0 \\ \end{matrix} \right.\ \ \]
\[x^{2} - 8x + 12 = 0\]
\[D_{1} = 16 - 12 = 4\]
\[x_{1} = 4 + 2 = 6;\ \ \]
\[x_{2} = 4 - 2 = 2.\]
\[x \in \lbrack 2;4\rbrack.\]
\[Ответ:x = 2;3;4.\]
\[\boxed{\text{321.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ \left( x^{2} + 6x \right)^{2} -\]
\[- 5 \cdot \left( x^{2} + 6x \right) = 24\]
\[Пусть\ t = x^{2} + 6x:\ \ \]
\[t^{2} - 5t = 24\]
\[t^{2} - 5t - 24 = 0\]
\[D = 25 + 4 \cdot 24 = 121\]
\[t_{1,2} = \frac{5 \pm 11}{2} = - 3;8;\]
\[1)\ x^{2} + 6x = - 3\]
\[x^{2} + 6x + 3 = 0\]
\[D_{1} = 9 - 3 = 6\]
\[x_{1,2} = - 3 \pm \sqrt{6}.\]
\[2)\ x^{2} + 6x = 8\]
\[x^{2} + 6x - 8 = 0\]
\[D_{1} = 9 + 8 = 17\]
\[x_{3,4} = - 3 \pm \sqrt{17}.\]
\[Ответ: - 3 \pm \sqrt{17};\ - 3 \pm \sqrt{6}.\]
\[\textbf{б)}\ \left( x^{2} - 2x - 5 \right)^{2} -\]
\[- 2 \cdot \left( x^{2} - 2x - 5 \right) = 3\]
\[Пусть\ t = x^{2} - 2x - 5:\ \]
\[t^{2} - 2t = 3\]
\[t^{2} - 2t - 3 = 0\]
\[D_{1} = 1 + 3 = 4\]
\[t_{1,2} = 1 \pm 2 = - 1;3.\]
\[1)\ x^{2} - 2x - 5 = - 1\]
\[x^{2} - 2x - 4 = 0\]
\[D_{1} = 1 + 4 = 5\]
\[x_{1,2} = 1 \pm \sqrt{5}.\]
\[2)\ x^{2} - 2x - 5 = 3\]
\[x^{2} - 2x - 8 = 0\]
\[D_{1} = 1 + 8 = 9\]
\[x_{1,2} = 1 \pm 3 = 4;\ - 2.\]
\[Ответ: - 2;4;1 \pm \sqrt{5}.\]
\[\textbf{в)}\ \left( x^{2} + 3x - 25 \right)^{2} -\]
\[- 2 \cdot \left( x^{2} + 3x - 25 \right) = - 7\]
\[Пусть\ x^{2} + 3x - 25 = t:\]
\[\ t^{2} - 2t + 7 = 0\]
\[D_{1} = 1 - 7 = - 6 < 0 \Longrightarrow\]
\[\Longrightarrow корней\ нет.\]
\[Ответ:\ корней\ нет.\]
\[\textbf{г)}\ (y + 2)^{4} - (y + 2)^{2} = 12\]
\[Пусть\ t = (y + 2)^{2};\ t \geq 0:\ \]
\[t^{2} - t - 12 = 0\]
\[D = 1 + 4 \cdot 12 = 49\]
\[t_{1,2} = \frac{1 \pm 7}{2} = 4; - 3.\]
\[так\ как\ \ t \geq 0:\]
\[(y + 2)^{2} = 4\]
\[y + 2 = \pm 2;\]
\[1)\ y + 2 = 2\ \ \]
\[y_{1} = 0.\]
\[2)\ y + 2 = - 2\ \ \]
\[y_{2} = - 4.\]
\[Ответ:\ - 4;0.\]
\[\textbf{д)}\ \left( x^{2} + 2x \right)\left( x^{2} + 2x + 2 \right) = 3\]
\[Пусть\ t = x^{2} + 2x:\ \]
\[t(t + 2) = 3\]
\[t^{2} + 2t - 3 = 0\]
\[D_{1} = 1 + 3 = 4\]
\[t_{1,2} = - 1 \pm 2 = - 3;1.\]
\[1)\ x^{2} + 2x = - 3\]
\[x^{2} + 2x + 3 = 0\]
\[D_{1} = 1 - 3 < 0 \Longrightarrow корней\ нет;\]
\[2)\ x^{2} + 2x = 1\]
\[x^{2} + 2x - 1 = 0\]
\[D_{1} = 1 + 1 = 2\]
\[x_{1,2} = - 1 \pm \sqrt{2}.\]
\[Ответ: - 1 \pm \sqrt{2}.\]
\[\textbf{е)}\ \left( x^{2} - x - 16 \right)\left( x^{2} - x + 2 \right) = 88\]
\[Пусть\ t = x^{2} - x:\ \]
\[(t - 16)(t + 2) = 88\]
\[t^{2} - 16t + 2t - 32 - 88 = 0\]
\[t^{2} - 14t - 120 = 0\]
\[D_{1} = 49 + 120 = 169\]
\[t_{1,2} = 7 \pm 13 = 20;\ - 6.\]
\[1)\ x^{2} - x = 20\]
\[x^{2} - x - 20 = 0\]
\[D = 1 + 4 \cdot 20 = 81\]
\[x_{1,2} = \frac{1 \pm 9}{2} = 5; - 4.\]
\[2)\ x^{2} - x = - 6\]
\[x^{2} - x + 6 = 0\]
\[D = 1 - 4 \cdot 6 < 0 \Longrightarrow\]
\[\Longrightarrow корней\ нет.\]
\[Ответ: - 4;5.\]
\[\textbf{ж)}\ \left( 2x^{2} + 7x - 8 \right) \cdot\]
\[\cdot \left( 2x^{2} + 7x - 3 \right) - 6 = 0\]
\[Пусть\ \ t = 2x^{2} + 7x:\]
\[(t - 8)(t - 3) - 6 = 0\]
\[t^{2} - 11t + 18 = 0\]
\[t_{1} + t_{2} = 11;\ \ \ \ t_{1} \cdot t_{2} = 18\]
\[t_{1} = 2;\ \ t_{2} = 9.\]
\[1)\ 2x^{2} + 7x = 2\]
\[2x^{2} + 7x - 2 = 0\]
\[D = 49 + 4 \cdot 2 \cdot 2 = 65\]
\[x_{1,2} = \frac{- 7 \pm \sqrt{65}}{4}.\]
\[2)\ 2x^{2} + 7x = 9\]
\[2x^{2} + 7x - 9 = 0\]
\[D = 49 + 4 \cdot 2 \cdot 9 = 121\]
\[x_{3,4} = \frac{- 7 \pm 11}{4} = 1;\ - 4,5.\]
\[Ответ:\ - 4,5;1;\ \frac{- 7 \pm \sqrt{65}}{4}.\]
