Решебник по алгебре 9 класс Макарычев Задание 386

 

\[\boxed{\text{386\ (386).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ (x + 1,2)(6 - x)(x - 4) > 0\]

\[(x + 1,2)(x - 4)(x - 6) < 0\]

\[x \in ( - \infty;\ - 1,2) \cup (4;6).\]

\[\textbf{б)}\ \left( \frac{1}{3} - x \right)(0,5 - x)\left( \frac{1}{7} - x \right) < 0\]

\[\left( x - \frac{1}{7} \right)\left( x - \frac{1}{3} \right)(x - 0,5) > 0\]

\[x \in \left( \frac{1}{7};\frac{1}{3} \right) \cup (0,5; + \infty).\]

\[\textbf{в)}\ (x + 0,6)(1,6 + x)(1,2 - x) > 0\]

\[(x + 1,6)(x + 0,6)(x - 1,2) < 0\]

\[x \in ( - \infty; - 1,6) \cup ( - 0,6;1,2).\]

\[\textbf{г)}\ (1,7 - x)(1,8 + x)(1,9 - x) < 0\]

\[(x + 1,8)(x - 1,7)(x - 1,9) < 0\]

\[x \in ( - \infty; - 1,8) \cup (1,7;1,9).\]

\[\boxed{\text{386.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \left\{ \begin{matrix} (x - 2)(y + 3) = 160 \\ y - x = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y = x + 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (x - 2)(x + 1 + 3) = 160 \\ \end{matrix} \right.\ \]

\[\Longrightarrow \left\{ \begin{matrix} y = x + 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (x - 2)(x + 4) = 160 \\ \end{matrix} \right.\ \]

\[(x - 2)(x + 4) = 160\]

\[x^{2} - 2x + 4x - 8 - 160 = 0\]

\[x^{2} + 2x - 168 = 0\]

\[D_{1} = 1 + 168 = 169\]

\[x_{1} = - 1 + 13 = 12;\ \ \ \]

\[\ x_{2} = - 1 - 13 = - 14.\]

\[1)\ \left\{ \begin{matrix} x_{1} = 12 \\ y_{1} = 13 \\ \end{matrix} \right.\ \]

\[2)\ \left\{ \begin{matrix} x_{2} = - 14 \\ y_{2} = - 13. \\ \end{matrix} \right.\ \]

\[\textbf{б)}\ \left\{ \begin{matrix} (x - 1)(y + 10) = 9 \\ x - y = 11\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = y + 11\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (y + 11 - 1)(y + 10) = 9 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = y + 11\ \ \ \ \\ (y + 10)^{2} = 9 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = y + 11\ \ \\ y + 10 = \pm 3 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y_{1} = - 7 \\ x_{1} = 4\ \ \ \ \\ \end{matrix} \right.\ \ \ \ или\ \ \left\{ \begin{matrix} y_{2} = - 13 \\ x_{2} = - 2\ \ \\ \end{matrix} \right.\ .\]

\[Ответ:а)\ (12;13);( - 14;\ - 13);\ \ \]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ б)\ (4;\ - 7);( - 2;\ - 13).\]