ГДЗ по алгебре 11 класс Никольский Параграф 11. Равносильность неравенств на множествах Задание 12

\[\boxed{\mathbf{12.}}\]

\[\textbf{а)}\ \sqrt{- x^{2} - 5x} > \sqrt{- x - 3}\]

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

\[x^{2} + 5x \leq 0\]

\[x(x + 5) \leq 0\]

\[- 5 \leq x \leq 0.\]

\[- x - 3 \geq 0\]

\[- x \geq 3\]

\[x \leq - 3.\]

\[M = \lbrack - 5; - 3\rbrack\text{.\ }\]

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

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

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

\[D_{1} = 4 + 3 = 7\]

\[x_{1} = - 2 + \sqrt{7};\]

\[x_{2} = - 2 - \sqrt{7}.\]

\[\sqrt{7} - 2 < x < - \sqrt{7} - 2.\]

\[x \in \left( - \sqrt{7} - 2; - 3 \right\rbrack.\]

\[Ответ:\ x \in \left( - \sqrt{7} - 2;\ - 3 \right\rbrack.\]

\[\textbf{б)}\ \sqrt{x^{2} - 6x} > \sqrt{- x - 1}\]

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

\[x(x - 6) \geq 0\]

\[x \leq 0;\ \ \ x \geq 6.\]

\[- x - 1 \geq 0\]

\[- x \geq 1\]

\[x \leq - 1.\]

\[M = ( - \infty; - 1\rbrack.\]

\[x^{2} - 6x > - x - 1\]

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

\[D = 25 - 4 = 21\]

\[x_{1} = \frac{5 + \sqrt{21}}{2};\]

\[x_{2} = \frac{5 - \sqrt{21}}{2};\]

\[\left( x - \frac{5 - \sqrt{21}}{2} \right)\left( x + \frac{5 + \sqrt{21}}{2} \right) > 0\]

\[x < \frac{5 - \sqrt{21}}{2};\]

\[x > \frac{5 + \sqrt{21}}{2}.\]

\[x \in ( - \infty; - 1\rbrack.\]

\[Ответ:\ x \in ( - \infty; - 1\rbrack.\]

\[\textbf{в)}\ \sqrt{x^{2} - x} > \sqrt{3x - 1}\]

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

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

\[x \leq 0;\ \ x \geq 1.\]

\[3x - 1 \geq 0\]

\[x \geq \frac{1}{3}.\]

\[M = \lbrack 1; + \infty).\]

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

\[x^{2} - 4x + 1 > 0\]

\[D_{1} = 4 - 1 = 3\]

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

\[x_{2} = 2 - \sqrt{3} < 1.\]

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

\[x < 2 - \sqrt{3};x > 2 + \sqrt{3}.\]

\[x \in \left( 2 + \sqrt{3}; + \infty \right).\]

\[Ответ:\ x \in \left( 2 + \sqrt{3}; + \infty \right).\]

\[\textbf{г)}\ \sqrt{x^{2} - 7x} > \sqrt{- x - 2}\]

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

\[x(x - 7) \geq 0\]

\[x \leq 0;\ \ x \geq 7.\]

\[- x - 2 \geq 0\]

\[- x \geq 2\]

\[x \leq - 2.\]

\[M = ( - \infty; - 2\rbrack.\]

\[x^{2} - 7x > - x - 2\]

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

\[D_{1} = 9 - 2 = 7\]

\[x_{1} = 3 + \sqrt{7} > - 2;\]

\[x_{2} = 3 - \sqrt{7} > - 2.\]

\[x < 3 - \sqrt{7};x > 3 + \sqrt{7}.\]

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

\[Ответ:\ x \in ( - \infty; - 2\rbrack.\]