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МатематикаРешебник по алгебре 9 класс Макарычев Задание 347
Задание 347
\[\boxed{\text{347\ (347).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ \left( x^{2} + 8x \right)^{2} -\]
\[- 4 \cdot (x + 4)^{2} = 256\]
\[x^{2} + 8x + 116 = (x + 4)^{2} \Longrightarrow\]
\[\Longrightarrow x^{2} + 8x = (x + 4)^{2} - 16.\]
\[Пусть\ \ x^{2} + 8x = a;\ \ \]
\[(x + 4)^{2} = a + 16:\]
\[\ a^{2} - 4 \cdot (a + 16) = 256\]
\[a^{2} - 4a - 64 - 256 = 0\]
\[a^{2} - 4a - 320 = 0\]
\[D_{1} = 2^{2} + 320 = 324\]
\[a_{1} = 2 + 18 = 20;\ \]
\[\ a_{2} = 2 - 18 = - 16.\]
\[1)\ x^{2} + 8x = 20\]
\[x^{2} + 8x - 20 = 0\]
\[D_{1} = 16 + 20 = 36\]
\[x_{1} = - 4 + 6 = 2;\ \]
\[\ x_{2} = - 4 - 6 = - 10.\]
\[2)\ x^{2} + 8x = - 16\]
\[x^{2} + 8x + 16 = 0\]
\[(x + 4)^{2} = 0\]
\[x + 4 = 0\]
\[x_{3} = - 4.\]
\[Ответ:x = - 10;x = - 4;x = 2.\]
\[\textbf{б)}\ 2 \cdot \left( x^{2} - 6x \right)^{2} -\]
\[- 120 \cdot (x - 3)^{2} = 8;\]
\[Пусть\ \ x^{2} - 6x = a;\ \ (x - 3)^{2} =\]
\[= x^{2} - 6x + 9 = a + 9:\]
\[2a^{2} - 120 \cdot (a + 9) = 8\ \ \ \ \ \ |\ :2\]
\[a^{2} - 60 \cdot (a + 9) - 4 = 0\]
\[a^{2} - 60a - 544 = 0\]
\[D_{1} = 30^{2} + 544 = 1444\]
\[a_{1} = 30 + 38 = 68;\ \ \]
\[a_{2} = 30 - 38 = - 8.\]
\[1)\ x^{2} - 6x = 68\]
\[x^{2} - 6x - 68 = 0\]
\[D_{1} = 9 + 68 = 77\]
\[x_{1,2} = 3 \pm \sqrt{77};\]
\[2)\ x^{2} - 6x = - 8\]
\[x^{2} - 6x + 8 = 0\]
\[D_{1} = 9 - 8 = 1\]
\[x_{1} = 3 + 1 = 4;\ \ x_{2} = 3 - 1 = 2.\]
\[Ответ:\ \ x = 2;x = 4;\]
\[x = 3 - \sqrt{77};x = 3 + \sqrt{77}\text{.\ }\]
\[\boxed{\text{347.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
Пояснение.
Решение.
\[\textbf{а)}\ \left\{ \begin{matrix} 4x^{2} - 27x - 7 > 0 \\ x > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[4x^{2} - 27x - 7 = 0\]
\[D = 27^{2} + 4 \cdot 4 \cdot 7 = 841\]
\[x_{1,2} = \frac{27 \pm 29}{8} = - \frac{1}{4};7.\]
\[\left\{ \begin{matrix} 4 \cdot \left( x + \frac{1}{4} \right)(x - 7) > 0 \\ x > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[x \in (7; + \infty).\]
\[\textbf{б)}\ \left\{ \begin{matrix} - 3x^{2} + 17x + 6 < 0 \\ x < 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \Longrightarrow \right.\ \]
\[\Longrightarrow \left\{ \begin{matrix} 3x^{2} - 17x - 6 > 0 \\ x < 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[3x^{2} - 17x - 6 = 0\]
\[D = 289 + 4 \cdot 3 \cdot 6 = 361\]
\[x_{1,2} = \frac{17 \pm 19}{6} = - \frac{1}{3};6.\]
\[\left\{ \begin{matrix} 3 \cdot \left( x + \frac{1}{3} \right)(x - 6) > 0 \\ x < 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[x \in \left( - \infty;\ - \frac{1}{3} \right).\]
\[\textbf{в)}\ \left\{ \begin{matrix} x + 1 < 0\ \ \ \ \ \\ 2x^{2} - 18 > 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x < - 1\ \ \ \ \ \ \\ x^{2} - 9 > 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x < - 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (x - 3)(x + 3) > 0 \\ \end{matrix} \right.\ \]
\[x \in ( - \infty;\ - 3).\]
\[\textbf{г)}\ \left\{ \begin{matrix} x - 4 > 0\ \ \ \ \ \ \ \ \\ 3x^{2} - 15x < 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x > 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3x(x - 5) < 0 \\ \end{matrix}\ \right.\ \ \]
\[x \in (4;5).\]
