Решебник по алгебре 9 класс Макарычев Задание 1015

\[\boxed{\text{1015\ (1015).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \left\{ \begin{matrix} x² - 7x + 6 \leq 0\ \ \ \\ x² - 8x + 15 \geq 0 \\ \end{matrix} \right.\ \]

\[1)\ x^{2} - 7x + 6 \leq 0\]

\[D = 49 - 24 = 25\]

\[x_{1} = \frac{7 + 5}{2} = 6,\ \ \]

\[x_{2} = \frac{7 - 5}{2} = 1,\]

\[x \in \lbrack 1;6\rbrack.\]

\[2)\ x² - 8x + 15 \geq 0\]

\[D = 64 - 60 = 4,\]

\[x_{1} = \frac{8 + 2}{2} = 5,\ \ \]

\[x_{2} = \frac{8 - 2}{2} = 3,\]

\[x \in ( - \infty;3\rbrack\lbrack 5; + \infty).\]

\[Ответ:\ x = 1;2;3;5;6.\]

\[\textbf{б)}\ \left\{ \begin{matrix} x² + 1 \geq 0\ \ \ \ \ \ \ \ \ \ \ \\ x² - 6x + 8 \leq 0 \\ \end{matrix} \right.\ \]

\[1)\ x² + 1 \geq 0\]

\[x^{2} \geq - 1 \Longrightarrow при\ любом\ x.\]

\[2)\ x² - 6x + 8 \leq 0\]

\[D = 36 - 32 = 4\]

\[x_{1} = \frac{6 + 2}{2} = 4,\ \ \]

\[x_{2} = \frac{6 - 2}{2} = 2,\]

\[x \in \lbrack 2;4\rbrack.\]

\[Ответ:x = 2;3;4.\]