Решебник по алгебре 11 класс Никольский Параграф 11. Равносильность неравенств на множествах Задание 24

\[\boxed{\mathbf{24.}}\]

\[\textbf{а)}\log_{25}{(x^{2} - 7)} > \log_{25}{(x - 1)}\]

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

\[x^{2} > 7\]

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

\[x - 1 > 0\]

\[x > 1.\]

\[M = \left( \sqrt{7}; + \infty \right).\]

\[x^{2} - 7 > x - 1\]

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

\[x_{1} + x_{2} = 1;\ \ x_{1} \cdot x_{2} = - 6\]

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

\[(x + 2)(x - 3) > 0\]

\[x < - 2;\ \ x > 3.\]

\[Решение\ неравенства:\]

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

\[Ответ:\ x \in (3; + \infty).\]

\[\textbf{б)}\log_{7}\left( x^{2} - 4 \right) > \log_{7}{(3x + 6)}\]

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

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

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

\[3x + 6 > 0\]

\[3x > - 6\]

\[x > - 2.\]

\[M = (2; + \infty).\]

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

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

\[x_{1} + x_{2} = 3;\ \ x_{1} \cdot x_{2} = - 10\]

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

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

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

\[Решение\ неравенства:\]

\[x \in (5; + \infty).\]

\[Ответ:\ x \in (5; + \infty).\]

\[\textbf{в)}\log_{\frac{1}{7}}\left( x^{2} - 3x \right) > \log_{\frac{1}{7}}(2x - 4)\]

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

\[x(x - 3) > 0\]

\[x < 0;\ \ x > 3.\]

\[2x - 4 > 0\]

\[2x > 4\]

\[x > 2.\]

\[M = (3; + \infty).\]

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

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

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

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

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

\[1 < x < 4.\]

\[Решение\ неравенства:\]

\[x \in (3;4).\]

\[Ответ:\ x \in (3;4).\]

\[\textbf{г)}\log_{\frac{1}{25}}\left( x^{2} - 4x \right) > \log_{\frac{1}{25}}{(2x - 5)}\]

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

\[x(x - 4) > 0\]

\[x < 0;\ \ x > 4.\]

\[2x - 5 > 0\]

\[2x > 5\]

\[x > 2,5.\]

\[M = (4; + \infty).\]

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

\[x^{2} - 6x + 5 < 0\]

\[D_{1} = 9 - 5 = 4\]

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

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

\[(x - 1)(x - 5) < 0\]

\[1 < x < 5.\]

\[Решение\ неравенства:\]

\[x \in (4;5).\]

\[Ответ:\ x \in (4;5).\]