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МатематикаГДЗ по алгебре 11 класс Никольский Параграф 11. Равносильность неравенств на множествах Задание 40
Задание 40
\[\boxed{\mathbf{40.}}\]
\[\textbf{а)}\lg(x + 1) + \lg(x - 8) >\]
\[> \lg{(2x - 8)}\]
\[x + 1 > 0\]
\[x > - 1.\]
\[x - 8 > 0\]
\[x > 8.\]
\[2x - 8 > 0\]
\[2x > 8\]
\[x > 4.\]
\[M = (8; + \infty).\]
\[\lg{(x + 1)(x - 8)} > \lg{(2x - 8)}\]
\[(x + 1)(x - 8) > 2x - 8\]
\[x^{2} + x - 8x - 8 - 2x + 8 > 0\]
\[x^{2} - 9x > 0\]
\[x(x - 9) > 0\]
\[x < 0;\ \ x > 9.\]
\[Решение\ неравенства:\]
\[x \in (9; + \infty).\]
\[Ответ:x \in (9; + \infty).\]
\[\textbf{б)}\log_{2}(x + 1) + \log_{2}(3x - 1) >\]
\[> \log_{2}{(9x + 5)}\]
\[x + 1 > 0\]
\[x > - 1.\]
\[3x - 1 > 0\]
\[3x > 1\]
\[x > \frac{1}{3}.\]
\[9x + 5 > 0\]
\[9x > - 5\]
\[x > - \frac{5}{9}.\]
\[M = \left( \frac{1}{3}; + \infty \right).\]
\[\log_{2}{(x + 1)(3x - 1)} >\]
\[> \log_{2}{(9x + 5)}\]
\[(x + 1)(3x - 1) > 9x + 5\]
\[3x^{2} + 3x - x - 1 - 9x - 5 > 0\]
\[3x^{2} - 7x - 6 > 0\]
\[D = 49 + 72 = 121\]
\[x_{1} = \frac{7 + 11}{6} = 3;\]
\[x_{2} = \frac{7 - 11}{6} = - \frac{4}{6} = - \frac{2}{3};\]
\[\left( x + \frac{2}{3} \right)(x - 3) > 0\]
\[x < - \frac{2}{3};\ \ x > 3.\]
\[Решение\ неравенства:\]
\[x \in (3; + \infty).\]
\[Ответ:x \in (3; + \infty).\]
\[3x + 1 > 0\]
\[3x > - 1\]
\[x > - \frac{1}{3}.\]
\[2x - 1 > 0\]
\[2x > 1\]
\[x > 0,5.\]
\[5x - 1 > 0\]
\[5x > 1\]
\[x > 0,2.\]
\[M = (0,5; + \infty).\]
\[\log_{\frac{1}{2}}{(3x + 1)(2x - 1)} <\]
\[< \log_{\frac{1}{2}}{(5x - 1)}\]
\[(3x + 1)(2x - 1) > 5x - 1\]
\[6x^{2} + 2x - 3x - 1 - 5x + 1 > 0\]
\[6x^{2} - 6x > 0\]
\[6x(x - 1) > 0\]
\[x < 0;\ \ x > 1.\]
\[Решение\ неравенства:\]
\[x \in (1; + \infty).\]
\[Ответ:x \in (1; + \infty).\]
\[4x + 1 > 0\]
\[4x > - 1\]
\[x > - 0,25.\]
\[2x - 1 > 0\]
\[2x > 1\]
\[x > 0,5.\]
\[10x + 7 > 0\]
\[10x > - 7\]
\[x > - 0,7.\]
\[M = (0,5; + \infty).\]
\[\log_{\frac{1}{3}}{(4x + 1)(2x - 1)} <\]
\[< \log_{\frac{1}{3}}{(10x + 7)}\]
\[(4x + 1)(2x - 1) > 10x + 7\]
\[8x^{2} + 2x - 4x - 1 - 10x - 7 >\]
\[> 0\]
\[8x^{2} - 12x - 8 > 0\ \ \ \ |\ :4\]
\[2x^{2} - 3x - 2 > 0\]
\[D = 9 + 16 = 25\]
\[x_{1} = \frac{3 + 5}{4} = 2;\]
\[x_{2} = \frac{3 - 5}{4} = - 0,5.\]
\[(x + 0,5)(x - 2) > 0\]
\[x < - 0,5;\ \ x > 2.\]
\[Решение\ неравенства:\]
\[x \in (2; + \infty).\]
\[Ответ:x \in (2; + \infty).\]