Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1400

\[\boxed{\mathbf{1400}\mathbf{.}}\]

\[1)\ |2x - 3| < x\]

\[2x - 3 \geq 0\]

\[2x \geq 3\]

\[x \geq 1,5.\]

\[x \geq 1,5:\]

\[2x - 3 < x\]

\[x < 3.\]

\[x < 1,5:\]

\[- (2x - 3) < x\]

\[- 2x + 3 < x\]

\[- 3x < - 3\]

\[x > 1.\]

\[Ответ:\ \ 1 < x < 3.\]

\[2)\ |4 - x| > x\]

\[4 - x \geq 0\]

\[x \leq 4.\]

\[x \leq 4:\]

\[4 - x > x\]

\[2x < 4\]

\[x < 2.\]

\[x > 4:\]

\[- (4 - x) > x\]

\[- 4 + x > x\]

\[0x < - 4\]

\[корней\ нет.\]

\[Ответ:\ \ x < 2.\]

\[3)\ \left| x^{2} - 7x + 12 \right| \leq 6\]

\[x^{2} - 7x + 12 \geq 0\]

\[D = 49 - 48 = 1\]

\[x_{1} = \frac{7 - 1}{2} = 3;\]

\[x_{2} = \frac{7 + 1}{2} = 4;\]

\[(x - 3)(x - 4) \geq 0\]

\[x \leq 3\ \ и\ \ x \geq 4.\]

\[x \leq 3\ и\ x \geq 4:\]

\[x^{2} - 7x + 12 \leq 6\]

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

\[D = 49 - 24 = 25\]

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

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

\[(x - 1)(x - 6) \leq 0\]

\[1 \leq x \leq 6.\]

\[3 < x < 4:\]

\[- \left( x^{2} - 7x + 12 \right) \leq 6\]

\[- x^{2} + 7x - 12 - 6 \leq 0\]

\[x^{2} - 7x + 18 \geq 0\]

\[при\ любом\ \text{x.}\]

\[D = 49 - 72 = - 23 < 0.\]

\[Ответ:\ \ 1 \leq x \leq 6.\]

\[4)\ \left| x^{2} - 3x - 4 \right| > 6\]

\[x^{2} - 3x - 4 \geq 0\]

\[D = 9 + 16 = 25\]

\[x_{1} = \frac{3 - 5}{2} = - 1;\]

\[x_{2} = \frac{3 + 5}{2} = 4;\]

\[(x + 1)(x - 4) \geq 0\]

\[x \leq - 1\ \ и\ \ x \geq 4.\]

\[x \leq - 1\ и\ x \geq 4:\]

\[x^{2} - 3x - 4 > 6\]

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

\[D = 9 + 40 = 49\]

\[x_{1} = \frac{3 - 7}{2} = - 2;\]

\[x_{2} = \frac{3 + 7}{2} = 5;\]

\[(x + 2)(x - 5) > 0\]

\[x < - 2\ \ и\ \ x > 5.\]

\[\ - 1 < x < 4:\]

\[- \left( x^{2} - 3x - 4 \right) > 6\]

\[- x^{2} + 3x + 4 - 6 > 0\]

\[x^{2} - 3x + 2 < 0\]

\[D = 9 - 8 = 1\]

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

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

\[(x - 1)(x - 2) < 0\]

\[1 < x < 2.\]

\[Ответ:\ \ \ \]

\[x < - 2;\ \ 1 < x < 2;\ \ x > 5.\]

\[5)\ \left| 2x^{2} - x - 1 \right| \geq 5\]

\[2x^{2} - x - 1 \geq 0\]

\[D = 1 + 8 = 9\]

\[x_{1} = \frac{1 - 3}{2 \bullet 2} = - 0,5;\]

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

\[(x + 0,5)(x - 1) \geq 0\]

\[x \leq - 0,5\ \ и\ \ x \geq 1.\]

\[x \leq - 0,5\ и\ x \geq 1:\]

\[2x^{2} - x - 1 \geq 5\]

\[2x^{2} - x - 6 \geq 0\]

\[D = 1 + 48 = 49\]

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

\[x_{2} = \frac{1 + 7}{2 \bullet 2} = 2;\]

\[(x + 1,5)(x - 2) \geq 0\]

\[x \leq - 1,5\ \ и\ \ x \geq 2.\]

\[- 0,5 < x < 1:\]

\[- \left( 2x^{2} - x - 1 \right) \geq 5\]

\[- 2x^{2} + x + 1 - 5 \geq 0\]

\[2x^{2} - x + 4 \leq 0\]

\[корней\ нет;\]

\[D = 1 - 32 = - 31 < 0.\]

\[Ответ:\ \ \ x \leq - 1,5;\ \ x \geq 2.\]

\[6)\ \left| 3x^{2} - x - 4 \right| < 2\]

\[3x^{2} - x - 4 \geq 0\]

\[D = 1 + 48 = 49\]

\[x_{1} = \frac{1 - 7}{2 \bullet 3} = - 1;\]

\[x_{2} = \frac{1 + 7}{2 \bullet 3} = \frac{4}{3};\]

\[(x + 1)\left( x - \frac{4}{3} \right) \geq 0\]

\[x \leq - 1\ \ и\ \ x \geq \frac{4}{3}.\]

\[x \leq - 1\ и\ x \geq \frac{4}{3}:\]

\[3x^{2} - x - 4 < 2\]

\[3x^{2} - x - 6 < 0\]

\[D = 1 + 71 = 73\]

\[x = \frac{1 \pm \sqrt{73}}{2 \bullet 3} = \frac{1 \pm \sqrt{73}}{6};\]

\[\left( x - \frac{1 - \sqrt{73}}{6} \right)\left( x - \frac{1 + \sqrt{73}}{6} \right) < 0\]

\[\frac{1 - \sqrt{73}}{6} < x < \frac{1 + \sqrt{73}}{6}.\]

\[- 1 < x < \frac{4}{3}:\]

\[- \left( 3x^{2} - x - 4 \right) < 2\]

\[- 3x^{2} + x + 4 - 2 < 0\]

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

\[D = 1 + 24 = 25\]

\[x_{1} = \frac{1 - 5}{2 \bullet 3} = - \frac{2}{3};\]

\[x_{2} = \frac{1 + 5}{2 \bullet 3} = 1;\]

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

\[x < - \frac{2}{3}\text{\ \ }и\ \ x > 1.\]

\[Ответ:\ \ \frac{1 - \sqrt{73}}{6} < x < - \frac{2}{3};\ \ \]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1 < x < \frac{1 + \sqrt{73}}{6}.\]