Решебник по алгебре 11 класс Никольский Параграф 9. Равносильность уравнений и неравенств системам Задание 61

\[\boxed{\mathbf{61.}}\]

\[|x| \leq \frac{3x^{2} + 6x - 24}{x^{2} + 2x - 8}\]

\[|x| \leq \frac{3 \cdot (x^{2} + 2x - 8)}{x^{2} + 2x - 8}\]

\[|x| \leq 3\]

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

\[ОДЗ:\]

\[x^{2} + 2x - 8 \neq 0\]

\[D_{1} = 1 + 8 = 9\]

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

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

\[x \neq - 4;\ \ x \neq 2.\]

\[x \in \lbrack - 3;2) \cup (2;3\rbrack.\]

\[Ответ:\ x \in \lbrack - 3;2) \cup (2;3\rbrack.\]

\[|x| \leq \frac{4x^{2} - 12x - 40}{x^{2} - 3x - 10}\]

\[|x| \leq \frac{4(x^{2} - 3x - 10)}{x^{2} - 3x - 10}\]

\[|x| \leq 4\]

\[- 4 \leq x \leq 4.\]

\[ОДЗ:\]

\[x^{2} - 3x - 10 \neq 0\]

\[x_{1} + x_{2} = 3;\ \ x_{1} \cdot x_{2} = - 10\]

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

\[x \neq - 2;\ \ x \neq 5.\]

\[x \in \lbrack - 4; - 2) \cup ( - 2;4\rbrack.\]

\[Ответ:\ x \in \lbrack - 4; - 2) \cup ( - 2;4\rbrack.\]