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МатематикаРешебник по алгебре 9 класс Макарычев Задание 387
Задание 387
\[\boxed{\text{387\ (387).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ (3x - 5)(x + 4)(2 - x) = 0\]
\[x_{1} = \frac{5}{3} = 1\frac{2}{3};\ \ x_{2} = - 4;\ \ x_{3} = 2.\ \]
\[\textbf{б)}\ (3x - 5)(x + 4)(2 - x) > 0\]
\[3 \cdot (x + 4)\left( x - \frac{5}{3} \right)(x - 2) < 0\]
\[x \in ( - \infty;\ - 4) \cup \left( 1\frac{2}{3};2 \right).\]
\[\textbf{в)}\ (3x - 5)(x + 4)(2 - x) < 0\]
\[3 \cdot (x + 4)\left( x - \frac{5}{3} \right)(x - 2) > 0\]
\[x \in \left( - 4;1\frac{2}{3} \right) \cup (2; + \infty).\]
\[\boxed{\text{387.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
Пояснение.
Решение.
\[\textbf{а)}\ \left\{ \begin{matrix} 6 \cdot (x - y) - 50 = y \\ y - xy = 24\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} 6y - 6x - 50 - y = 0 \\ y - xy = 24\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} 5y - 6x - 50 = 0 \\ y - xy = 24\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} 5y = 6x + 50 \\ y - xy = 24\ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = 1,2x + 10\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 1,2x + 10 - 1,2x^{2} - 10x = 24 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = 1,2x + 10\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 1,2x^{2} + 8,8x + 14 = 0 \\ \end{matrix} \right.\ \]
\[1,2x^{2} + 8,8x + 14 = 0\ \ \ \ | \cdot 5\]
\[6x^{2} + 44x + 70 = 0\ \ \ \ \ \ \ \ \ |\ :2\]
\[3x^{2} + 22x + 35 = 0\]
\[D_{1} = 11^{2} - 3 \cdot 35 =\]
\[= 121 - 105 = 16\]
\[x_{1,2} = \frac{- 11 \pm 4}{3} = - 5;\ - 2\frac{1}{3}.\]
\[1)\ \left\{ \begin{matrix} x_{1} = - 5 \\ y_{1} = 4\ \ \\ \end{matrix} \right.\ \]
\[2)\ \left\{ \begin{matrix} x_{2} = 7\frac{1}{5}\text{\ \ \ \ } \\ y_{2} = - 2\frac{1}{3} \\ \end{matrix} \right.\ .\]
\[\textbf{б)}\ \left\{ \begin{matrix} p + 5t = 2 \cdot (p + t) \\ pt - t = 10\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} p + 5t = 2p + 2t \\ pt - t = 10\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} p = 3t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3t^{2} - t - 10 = 0 \\ \end{matrix} \right.\ \]
\[3t^{2} - t - 10 = 0\]
\[D = 1 + 4 \cdot 3 \cdot 10 = 121\]
\[t_{1,2} = \frac{1 \pm 11}{6} = 2;\ - 1\frac{2}{3}.\]
\[1)\ \left\{ \begin{matrix} t_{1} = 2 \\ p_{1} = 6 \\ \end{matrix} \right.\ \text{\ \ }\]
\[2)\ \left\{ \begin{matrix} t_{2} = - 1\frac{2}{3} \\ p_{2} = - 5\ \ \\ \end{matrix} \right.\ .\]
\[Ответ:а)\ ( - 5;4);\left( - 2\frac{1}{3};7\frac{1}{5} \right);\ \ \]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ б)\ (6;2);\ \ \left( - 5;\ - 1\frac{2}{3} \right).\]
