Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1579

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\[|x - 5a| \leq 4a - 3\]

\[Имеет\ решения\ при:\]

\[4a - 3 \geq 0\]

\[4a \geq 3\]

\[a \geq \frac{3}{4}.\]

\[a = \frac{3}{4}:\]

\[x - 5a = 0\]

\[x = 5a = 5 \bullet \frac{3}{4} = \frac{15}{4}.\]

\[a > \frac{3}{4}:\]

\[x - 5a \geq 0\]

\[x \geq 5a.\]

\[x \geq 5a:\]

\[x - 5a \leq 4a - 3\]

\[x \leq 9a - 3.\]

\[x < 5a:\]

\[- (x - 5a) \leq 4a - 3\]

\[- x + 5a \leq 4a - 3\]

\[- x \leq - a - 3\]

\[x \geq a + 3.\]

\[если\ a < \frac{3}{4},\ то\ решений\ нет;\ \ \]

\[если\ a = \frac{3}{4},\ то\ x = \frac{15}{4};\ \ \]

\[если\ a > \frac{3}{4},\ то\ a + 3 \leq x \leq 9a - 3.\]

\[Решения\ совпадают\ с\ \]

\[решениями\ неравенства:\]

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

\[D = 16 + 20 = 36\]

\[x_{1} = \frac{4 - 6}{2} = - 1;\]

\[x_{2} = \frac{4 + 6}{2} = 5;\]

\[(x + 1)(x - 5) < 0\]

\[- 1 < x < 5.\]

\[\left\{ \begin{matrix} a + 3 > - 1 \\ 9a - 3 < 5\ \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} a > - 4 \\ 9a < 8\ \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} a > - 4 \\ a < \frac{8}{9}\text{\ \ \ } \\ \end{matrix} \right.\ \]

\[Ответ:\ \ \frac{3}{4} \leq a < \frac{8}{9}.\]