Алгебра
Английский
Биология
География
Геометрия
Русский
История
Обществознание
Физика
Химия
Информатика
Литература
МатематикаГДЗ по алгебре 11 класс Никольский Параграф 11. Равносильность неравенств на множествах Задание 18
Задание 18
\[\boxed{\mathbf{18.}}\]
\[\textbf{а)}\ \frac{x^{2}}{\sqrt{x^{2} - 16}} > \frac{5x - 5}{\sqrt{x^{2} - 16}}\]
\[x^{2} - 16 > 0\]
\[(x + 4)(x - 4) > 0\]
\[x < - 4;\ \ x > 4.\]
\[M = ( - \infty; - 4) \cup (4; + \infty).\]
\[x^{2} > 5x - 5\]
\[x^{2} - 5x + 5 > 0\]
\[D = 25 - 20 = 5\]
\[x_{1} = \frac{5 + \sqrt{5}}{2};\]
\[x_{2} = \frac{5 - \sqrt{5}}{2};\]
\[x < \frac{5 - \sqrt{5}}{2};x > \frac{5 + \sqrt{5}}{2}.\]
\[Ответ:x \in ( - \infty; - 4) \cup (4; + \infty).\]
\[\textbf{б)}\ \frac{x^{2}}{\sqrt{5x - x^{2}}} > \frac{7x - 3}{\sqrt{5x - x^{2}}}\]
\[5x - x^{2} > 0\]
\[x^{2} - 5x < 0\]
\[x(x - 5) < 0\]
\[0 < x < 5.\]
\[M = (0;5).\]
\[x^{2} > 7x - 3\]
\[x^{2} - 7x + 3 > 0\]
\[D = 49 - 12 = 37\]
\[x_{1} = \frac{7 + \sqrt{37}}{2} > 6;\]
\[x_{2} = \frac{7 - \sqrt{37}}{2} > 0.\]
\[x < \frac{7 - \sqrt{37}}{2};x > \frac{7 + \sqrt{37}}{2}.\]
\[Ответ:x \in \left( 0;\ \frac{7 - \sqrt{37}}{2} \right).\]
\[\textbf{в)}\ \frac{x^{2}}{\sqrt{3x - x^{2} - 2}} < \frac{3 - 6x}{\sqrt{3x - x^{2} - 2}}\]
\[3x - x^{2} - 2 > 0\]
\[x^{2} - 3x + 2 < 0\]
\[x_{1} + x_{2} = 3;\ \ x_{1} \cdot x_{2} = 2\]
\[x_{1} = 1;\ \ x_{2} = 2;\]
\[(x - 1)(x - 2) < 0\]
\[1 < x < 2.\]
\[M = (1;2).\]
\[x^{2} < 3 - 6x\]
\[x^{2} + 6x - 3 < 0\]
\[D_{1} = 9 + 3 = 12 = 2\sqrt{3}\]
\[x_{1} = - 3 + 2\sqrt{3} < 1;\]
\[x_{2} = - 3 - 2\sqrt{3} < 1.\]
\[\left( x + 3 - 2\sqrt{3} \right)\left( x + 3 + 2\sqrt{3} \right) < 0\]
\[- 3 - 2\sqrt{3} < x < - 3 + 2\sqrt{3}.\]
\[Ответ:нет\ корней.\]
\[\textbf{г)}\ \frac{x^{2}}{\sqrt{5 - 4x - x^{2}}} < \frac{11 - 4x}{\sqrt{5 - 4x - x^{2}}}\]
\[5 - 4x - x^{2} > 0\]
\[x^{2} + 4x - 5 < 0\]
\[D_{1} = 4 + 5 = 9\]
\[x_{1} = - 2 + 3 = 1;\]
\[x_{2} = - 2 - 3 = - 5;\]
\[(x + 5)(x - 1) < 0\]
\[- 5 < x < 1.\]
\[M = ( - 5;1).\]
\[x^{2} < 11 - 4x\]
\[x^{2} + 4x - 11 < 0\]
\[D_{1} = 4 + 11 = 15\]
\[x_{1} = - 2 + \sqrt{15} > 1;\]
\[x_{2} = - 2 - \sqrt{15} < - 5;\]
\[- 2 - \sqrt{15} < x < - 2 + \sqrt{15}.\]
\[Ответ:x \in ( - 5;1).\]