ГДЗ по алгебре 11 класс Никольский Параграф 11. Равносильность неравенств на множествах Задание 36

\[\boxed{\mathbf{36.}}\]

\[\textbf{а)}\ \left( \frac{1}{25} \right)^{\frac{(2x - 5)}{2}} > \left( \frac{1}{5} \right)^{\sqrt{6x^{2} - 31x + 25}}\]

\[6x^{2} - 31x + 25 \geq 0\]

\[D = 961 - 600 = 361 = 19^{2}\]

\[x_{1} = \frac{31 + 19}{12} = \frac{50}{12} = \frac{25}{6};\]

\[x_{2} = \frac{31 - 19}{12} = 1;\]

\[(x - 1)\left( x - \frac{25}{6} \right) \geq 0\]

\[x \leq 1;\ \ x \geq \frac{25}{6};\]

\[M = ( - \infty;1\rbrack \cup \left\lbrack \frac{25}{6}; + \infty \right).\]

\[\left( \frac{1}{5} \right)^{\frac{2(2x - 5)}{2}} > \left( \frac{1}{5} \right)^{\sqrt{6x^{2} - 31x + 25}}\]

\[\left( \frac{1}{5} \right)^{2x - 5} > \left( \frac{1}{5} \right)^{\sqrt{6x^{2} - 31x + 25}}\]

\[(5)^{- 2x + 5} > (5)^{- \sqrt{6x^{2} - 31x + 25}}\]

\[2x - 5 > \sqrt{6x^{2} - 31x + 25}\]

\[(2x - 5)^{2} > 6x^{2} - 31x + 25\]

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

\(> 6x^{2} - 31x + 25\)

\[Решение\ неравенства:\]

\[x \in ( - \infty;1\rbrack \cup \lbrack 5,5; + \infty).\]

\[Ответ:\ x \in ( - \infty;1\rbrack \cup \lbrack 5,5; + \infty).\]

\[\textbf{б)}\ \left( \frac{1}{36} \right)^{x - 3} > \left( \frac{1}{6} \right)^{\sqrt{5x^{2} - 41x + 36}}\]

\[5x^{2} - 41x + 36 \geq 0\]

\[D = 1681 - 720 = 961 = 31^{2}\]

\[x_{1} = \frac{41 + 31}{10} = 7,2;\]

\[x_{2} = \frac{41 - 31}{10} = 1;\]

\[(x - 1)(x - 7,2) \geq 0\]

\[x \leq 1;\ \ x \geq 7,2.\]

\[M = ( - \infty;1\rbrack \cup \lbrack 7,2; + \infty).\]

\[6^{- 2(x - 3)} > 6^{- \sqrt{5x^{2} - 41x + 36}}\]

\[\sqrt{5x^{2} - 41x + 36} > 2x - 6\]

\[5x^{2} - 41x + 36 > (2x - 6)^{2}\]

\[5x^{2} - 41x + 36 >\]

\[> 4x^{2} - 24x + 36\]

\[Решение\ неравенства:\]

\[x \in ( - \infty;1\rbrack \cup \lbrack 17; + \infty).\]

\(Ответ:\ x \in ( - \infty;1\rbrack \cup \lbrack 17; + \infty).\)