Решебник по алгебре 9 класс Макарычев Задание 243

 

\[\boxed{\text{243}\text{\ (243)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ y = x^{2} + 2x - 15\]

\[x_{b} = - \frac{b}{2a} = - \frac{2}{2} = - 1;\]

\[y_{b} = ( - 1)^{2} - 2 \cdot 1 - 15 = - 16;\]

\[x^{2} + 2x - 15 = 0\]

\[D_{1} = 1 + 15 = 16\]

\[x_{1} = - 1 - 4 = - 5;\ \ x_{2} =\]

\[= - 1 + 4 = 3\]

\[E(y) = \ \lbrack - 16;\ + \infty).\]

\[При\ x \in ( - \infty; - 1)\ функция\ \]

\[убывает\ и\ при\ \]

\[x \in ( - 1; + \infty)\ возрастает.\]

\[y < 0\ \ при\ x \in ( - 5;3)\ и\ y > 0\ \ \]

\[при\ \ x \in ( - \infty; - 5) \cup (3; + \infty).\]

\[\textbf{б)}\ y = 0,5x^{2} - 3x + 4\]

\[x_{b} = - \frac{b}{2a} = - \frac{- 3}{- 2 \cdot 0,5} = 3;\]

\[y_{b} = 0,5 \cdot 3^{2} - 3 \cdot 3 + 4 = - 0,5.\]

\[0,5x^{2} - 3x + 4 = 0\]

\[D = 9 - 0,5 \cdot 16 = 1\]

\[x_{1} = \frac{3 + 1}{1} = 4;x_{2} = \frac{3 - 1}{1} = 2.\]

\[E(y) = \lbrack - 0,5;\ + \infty).\]

\[При\ x \in ( - \infty;3)\ функция\ \]

\[убывает\ и\ при\ \]

\[x \in (3; + \infty)\ возрастает.\]

\[y < 0\ \ при\ x \in (2;4)\ и\ y > 0\ \ \]

\[при\ \ x \in ( - \infty;2) \cup (4; + \infty).\]

\(в)\ y = 4 - 0,5x^{2}\)

\[4 - 0,5x^{2} = 0\]

\[0,5x^{2} = 4\]

\[x^{2} = 8\]

\[x \approx \pm 2,8.\]

\[x_{b} = \frac{0}{1} = 0;\]

\[y_{b} = 4.\]

\[E(y) = ( - \infty;4\rbrack;\]

\[D(y) = R.\]

\[Функция\ возрастает\ при\ \]

\[x \in ( - \infty;0);\]

\(убывает\ при\ x \in (0; + \infty).\)

\[y > 0\ при\ - 2,8 < x < 2,8;\]

\[y < 0\ \ при\ x < - 2,8;\ \ x > 2,8.\]

\[\textbf{г)}\ y = 6x - 2x^{2}\]

\[x_{b} = - \frac{b}{2a} = - \frac{6}{- 4} = \frac{3}{2};\]

\[y_{b} = 6 \cdot \frac{3}{2} - 2 \cdot \frac{9}{4} = 4,5;\]

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

\[2x(3 - x) = 0\]

\[x_{1} = 0\ \ и\ \ \ x_{2} = 3;\]

\[E(y) = ( - \infty;4,5\rbrack\]

\[y > 0\ \ при\ \ x \in (0;\ 3)\ и\ \ \]

\[y < 0\ \ \ при\ \ \]

\[x \in ( - \infty;0) \cup (3; + \infty);\]

\[Функция\ возрастает\ \]

\[при\ x \in ( - \infty;1,5)\ \ и\ \]

\[убывает\ \ (1,5; + \infty).\]

\[\textbf{д)}\ y = (2x - 7)(x + 1)\]

\[y = 2x^{2} - 5x - 7\]

\[x_{b} = - \frac{- 5}{4} = \frac{5}{4} = 1,25;\]

\[y_{b} = 2 \cdot \frac{25}{4} - 5 \cdot \frac{5}{4} - 7 =\]

\[= - 10,125;\]

\[x_{1} = - 1\ \ \ и\ \ \ x_{2} = 3,5 \Longrightarrow\]

\[\Longrightarrow нули\ функции;\]

\[E(y) = \lbrack - 10,125;\ + \infty).\]

\[При\ x \in ( - \infty;1,25)\ функция\ \]

\[убывает\ и\ при\ \]

\[x \in (1,25; + \infty)\ возрастает.\]

\[y < 0\ \ при\ x \in ( - 1;3,5)\ и\ \]

\[y > 0\ \ при\ \ \]

\[x \in ( - \infty; - 1) \cup (3,5; + \infty).\]

\[\textbf{е)}\ y = (2 - x)(x + 6)\]

\[y = - x^{2} - 4x + 12\]

\[x_{b} = - \frac{- 4}{- 2} = - 2;\ \ y_{b} = 16;\]

\[x_{1} = 2\ \ и\ \ x_{2} = - 2 \Longrightarrow\]

\[\Longrightarrow нули\ функции;\]

\[E(y) = ( - \infty;16\rbrack\]

\[y > 0\ \ при\ \ x \in ( - 6;\ - 2)\ и\ \ \]

\[y < 0\ \ \ при\ \ \]

\[x \in ( - \infty; - 6) \cup (2; + \infty);\]

\[Функция\ возрастает\ \]

\[при\ x \in ( - \infty; - 2)\ \ и\ \]

\[убывает\ \ ( - 2; + \infty).\]

\[\boxed{\text{243.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ \left( \frac{x + 2}{x - 4} \right)^{2} + 16 \cdot \left( \frac{x - 4}{x + 2} \right)^{2} =\]

\[= 17\]

\[Пусть\ t = \left( \frac{x + 2}{x - 4} \right)^{2},\ \]

\[t + \frac{16}{t} - 17 = 0\]

\[t^{2} - 17t + 16 = 0,\ \ t \neq 0,\]

\[t_{1} = 1,\ \ t_{2} = 16;\]

\[1)\ при\ \ t_{1} = 1 \Longrightarrow \left( \frac{x + 2}{x - 4} \right)^{2} = 1;\]

\[\left\{ \begin{matrix} \frac{x + 2}{x - 4} = 1\ \ \\ \frac{x + 2}{x - 4} = - 1 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x + 2 = x - 4\ \ \ \\ x + 2 = - x + 4 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} 2 = - 4 \\ 2x = 2 \\ \end{matrix} \right.\ \Longrightarrow x = 1.\]

\[2)\ при\ \ t_{2} = 16 \Longrightarrow\]

\[\Longrightarrow \left( \frac{x + 2}{x - 4} \right)^{2} = 16:\]

\[\left\{ \begin{matrix} \frac{x + 2}{x - 4} = 4\ \ \ \ \\ \frac{x + 2}{x - 4} = - 4 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x + 2 = 4x - 16\ \ \ \\ x + 2 = - 4x + 16 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} 3x = 18 \\ 5x = 14 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow x_{1} = 6,\ \ x_{2} = 2,8;\]

\[Ответ:1;2,8;6.\]

\[\textbf{б)}\ \left( \frac{x + 1}{x - 3} \right)^{2} + 18 \cdot \left( \frac{x - 3}{x + 1} \right)^{2} =\]

\[= 11\]

\[Пусть\ \ t = \left( \frac{x + 1}{x - 3} \right)^{2},\]

\[t + \frac{18}{t} = 11,\]

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

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

\[1)\ при\ t_{1} = 2:\ \ \left( \frac{x + 1}{x - 3} \right)^{2} = 2 \Longrightarrow\]

\[\left\{ \begin{matrix} \frac{x + 1}{x - 3} = \sqrt{2\ \ } \\ \frac{x + 1}{x - 3} = - \sqrt{2} \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x + 1 = \sqrt{2x} - 3\sqrt{2}\text{\ \ } \\ x + 1 = - \sqrt{2x} + 3\sqrt{2} \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 7 + 4\sqrt{2} \\ x = 7 - 4\sqrt{2} \\ \end{matrix} \right.\ ,\]

\[2)\ при\ t_{2} = 9:\ \ \]

\[\left\{ \begin{matrix} \frac{x + 1}{x - 3} = 3 \\ \frac{x + 1}{x - 3} = - 3 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x + 1 = 3x - 9\ \ \ \ \\ x + 1 = - 3x + 9 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 5 \\ x = 2 \\ \end{matrix} \right.\ .\]

\[Ответ:\ \ 2;5;7 + 4\sqrt{2};7 - 4\sqrt{2}.\]