ГДЗ по алгебре 9 класс Макарычев 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.

\[\textbf{а)}\ \left\{ \begin{matrix} 4x^{2} - 27x - 7 > 0 \\ x > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[4x^{2} - 27x - 7 = 0\]

\[D = 27^{2} + 4 \cdot 4 \cdot 7 = 841\]

\[x_{1,2} = \frac{27 \pm 29}{8} = - \frac{1}{4};7.\]

\[\left\{ \begin{matrix} 4 \cdot \left( x + \frac{1}{4} \right)(x - 7) > 0 \\ x > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[x \in (7; + \infty).\]

\[\textbf{б)}\ \left\{ \begin{matrix} - 3x^{2} + 17x + 6 < 0 \\ x < 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \Longrightarrow \right.\ \]

\[\Longrightarrow \left\{ \begin{matrix} 3x^{2} - 17x - 6 > 0 \\ x < 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[3x^{2} - 17x - 6 = 0\]

\[D = 289 + 4 \cdot 3 \cdot 6 = 361\]

\[x_{1,2} = \frac{17 \pm 19}{6} = - \frac{1}{3};6.\]

\[\left\{ \begin{matrix} 3 \cdot \left( x + \frac{1}{3} \right)(x - 6) > 0 \\ x < 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[x \in \left( - \infty;\ - \frac{1}{3} \right).\]

\[\textbf{в)}\ \left\{ \begin{matrix} x + 1 < 0\ \ \ \ \ \\ 2x^{2} - 18 > 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x < - 1\ \ \ \ \ \ \\ x^{2} - 9 > 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x < - 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (x - 3)(x + 3) > 0 \\ \end{matrix} \right.\ \]

\[x \in ( - \infty;\ - 3).\]

\[\textbf{г)}\ \left\{ \begin{matrix} x - 4 > 0\ \ \ \ \ \ \ \ \\ 3x^{2} - 15x < 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x > 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3x(x - 5) < 0 \\ \end{matrix}\ \right.\ \ \]

\[x \in (4;5).\]