Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1394

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\[1)\ \frac{3x - 15}{x^{2} + 5x - 14} \geq 0\]

\[x^{2} + 5x - 14 = 0\ \]

\[D = 25 + 56 = 81\]

\[x_{1} = \frac{- 5 - 9}{2} = - 7;\]

\[x_{2} = \frac{- 5 + 9}{2} = 2;\]

\[(x + 7)(x - 2) = 0.\]

\[\frac{3(x - 5)}{(x + 7)(x - 2)} \geq 0\]

\[(x + 7)(x - 2)(x - 5) \geq 0\]

\[- 7 < x < 2\ \ и\ \ x \geq 5.\]

\[Ответ:\ \ x \in ( - 7;\ 2) \cup \lbrack 5;\ + \infty)\text{.\ }\]

\[2)\ \frac{x - 1}{x^{2} + 4x + 2} < 0\]

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

\[D = 16 - 8 = 8 = 4 \bullet 2\]

\[x = \frac{- 4 \pm \sqrt{8}}{2} = \frac{- 4 \pm 2\sqrt{2}}{2} =\]

\[= - 2 \pm \sqrt{2}.\]

\[\left( x - \left( - 2 - \sqrt{2} \right) \right)\left( x - \left( - 2 + \sqrt{2} \right) \right) = 0.\]

\[\frac{x - 1}{\left( x - \left( - 2 - \sqrt{2} \right) \right)\left( x - \left( - 2 + \sqrt{2} \right) \right)} < 0\]

\[\left( x - \left( - 2 - \sqrt{2} \right) \right)\left( x - \left( - 2 + \sqrt{2} \right) \right)(x - 1) < 0\]

\[x < - 2 - \sqrt{2};\]

\[- 2 + \sqrt{2} < x < 1.\]

\[Ответ:\ \ \]

\[x \in \left( - \infty;\ - 2 - \sqrt{2} \right) \cup \left( - 2 + \sqrt{2};\ 1 \right).\]

\[3)\ \frac{x^{2} + 2x - 8}{x^{2} - 2x - 3} > 0\]

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

\[D = 4 + 32 = 36:\]

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

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

\[(x + 4)(x - 2) = 0.\]

\[x^{2} - 2x - 3 = 0\]

\[D = 4 + 12 = 16\]

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

\[x_{2} = \frac{2 + 4}{2} = 3;\]

\[(x + 1)(x - 3) = 0\]

\[\frac{(x + 4)(x - 2)}{(x + 1)(x - 3)} > 0\]

\[(x + 4)(x + 1)(x - 2)(x - 3) > 0\]

\[x < - 4;\]

\[\ - 1 < x < 2;\]

\[x > 3.\]

\[Ответ:\ \ \]

\[x \in ( - \infty;\ - 4) \cup ( - 1;\ 2) \cup (3;\ + \infty).\]