Решебник по алгебре 11 класс Никольский Параграф 11. Равносильность неравенств на множествах Задание 10

\[\boxed{\mathbf{10.}}\]

\[\textbf{а)}\ \sqrt{3 - x} > x - 1\]

\[3 - x \geq 0\]

\[x \leq 3.\]

\[M = ( - \infty;3\rbrack.\]

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

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

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

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

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

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

\[- 1 < x < 2.\]

\[x \in ( - \infty;2).\]

\[Ответ:\ x \in ( - \infty;2).\]

\[\textbf{б)}\ \sqrt{6 - x} > 3x - 4\]

\[6 - x \geq 0\]

\[x \leq 6.\]

\[M = ( - \infty;6\rbrack.\]

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

\[6 - x > 9x^{2} - 24x + 16\]

\[9x^{2} - 23x + 10 < 0\]

\[D = 529 - 360 = 169\]

\[x_{1} = \frac{23 + 13}{18} = 2;\]

\[x_{2} = \frac{23 - 13}{18} = \frac{10}{18} = \frac{5}{9};\]

\[\left( x - \frac{5}{9} \right)(x - 2) < 0\]

\[\frac{5}{9} < x < 2.\]

\[x \in ( - \infty;2).\]

\[Ответ:\ x \in ( - \infty;2).\]

\[\textbf{в)}\ \sqrt{5 - x} > x - 3\]

\[5 - x \geq 0\]

\[x \leq 5.\]

\[M = ( - \infty;5\rbrack.\]

\[5 - x > (x - 3)^{2}\]

\[5 - x > x^{2} - 6x + 9\]

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

\[x_{1} + x_{2} = 5;\ \ x_{1} \cdot x_{5} = 4\]

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

\[(x - 1)(x - 4) < 0\]

\[1 < x < 4.\]

\[x \in ( - \infty;4).\]

\[Ответ:\ x \in ( - \infty;4).\]

\[\textbf{г)}\ \sqrt{4 - x} > 2x - 5\]

\[4 - x \geq 0\]

\[x \leq 4.\]

\[M = ( - \infty;4\rbrack.\]

\[4 - x > (2x - 5)^{2}\]

\[4 - x > 4x^{2} - 20x + 25\]

\[4x^{2} - 19x + 21 < 0\]

\[D = 361 - 336 = 25\]

\[x_{1} = \frac{19 + 5}{8} = 3;\]

\[x_{2} = \frac{19 - 5}{8} = \frac{14}{8} = \frac{7}{4};\]

\[\left( x - \frac{7}{4} \right)(x - 3) < 0\]

\[\frac{7}{4} < x < 3.\]

\[x \in ( - \infty;3).\]

\[Ответ:\ x \in ( - \infty;3).\]