ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 894

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

\[2x - 3 > - x\]

\[3x > 3\]

\[x > 1.\ \]

\[2x - 3 < x\]

\[2x - x < 3\]

\[x < 3.\]

\[Ответ:\ \ x \in (1;\ 3).\]

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

\[4 - x < - x\]

\[0x > 4\]

\[x \in \varnothing.\]

\[4 - x > x\]

\[2x < 4\]

\[x < 2.\]

\[Ответ:\ \ x \in ( - \infty;\ 2).\]

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

\[1)\ x^{2} - 7x + 12 \geq - 6\]

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

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

\[x \in R.\]

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

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

\[D = 49 - 24 = 25\]

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

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

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

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

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

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

\[1)\ x^{2} - 3x - 4 < - 6\]

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

\[D = 9 - 8 = 1\]

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

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

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

\[1 < x < 2.\]

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

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

\[D = 9 + 40 = 49\]

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

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

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

\[x < - 2;\ \text{\ \ }x > 5.\]

\[Ответ:\ \ \]

\[x \in ( - \infty;\ - 2) \cup (1;\ 2) \cup (5;\ + \infty).\]

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

\[1)\ 2x^{2} - x - 1 \leq - 5\]

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

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

\[x \in \varnothing.\]

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

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

\[D = 1 + 48 = 49\]

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

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

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

\[x \leq - 1,5;\ \text{\ \ }x \geq 2.\]

\[Ответ:\ \]

\[x \in ( - \infty;\ - 1,5\rbrack \cup \lbrack 2;\ + \infty).\]

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

\[1)\ 3x^{2} - x - 4 > - 2\]

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

\[D = 1 + 24 = 25\]

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

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

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

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

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

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

\[D = 1 + 71 = 73\]

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

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

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

\[Ответ:\ \ \]

\[x \in \left( \frac{1 - \sqrt{73}}{6};\ - \frac{2}{3} \right) \cup \left( 1;\ \frac{1 + \sqrt{73}}{6} \right).\]