50 решите неравенства 1. $2^{x^{2} -6x+0,5}\leq (16\sqrt{2} )^{-1}$ 2. $\frac{7}{9^{x}-2 } \geq \frac{2}{3^{x} -1}$

Ответы

света8612 01.10.2020 13:06

$1) \ 2^{x^{2} - 6x + 0,5} \leqslant (16\sqrt{2} )^{-1}\\2^{x^{2} - 6x + 0,5} \leqslant (2^{4,5})^{-1}\\2^{x^{2} - 6x + 0,5} \leqslant 2^{-4,5}\\x^{2} - 6x + 0,5\leqslant -4,5\\x^{x} - 6x + 5 \leqslant 0\\x^{x} - 6x + 5 = 0\\x_{1} = 1; \ \ \ x_{2} = 5\\x \in [1; \ 5]$

ответ: $x \in [1; \ 5]$

$2) \ \dfrac{7}{9^{x} - 2} \geqslant \dfrac{2}{3^{x} - 1}$

ОДЗ: $\left \{ {\bigg{9^{x} - 2 \neq 0} \atop \bigg{3^{x} - 1 \neq 0}} \right. \ \ \ \ \ \ \ \ \ \ \ \left \{ {\bigg{9^{x} \neq 2} \atop \bigg{3^{x} \neq 1}} \right. \ \ \ \ \ \ \ \ \ \ \left \{ {\bigg{x \neq \log_{9}2} \atop \bigg{x \neq 0 \ \ \ \ \ \ }} \right.$

$\dfrac{7}{3^{2x} - 2} \geqslant \dfrac{2}{3^{x} - 1}$

Замена: $3^{x} = t, \ t 0$

$\dfrac{7}{t^{2} - 2} \geqslant \dfrac{2}{t - 1}\\\dfrac{7}{t^{2} - 2} - \dfrac{2}{t - 1} \geqslant 0\\$

$\dfrac{7(t-1) - 2(t^{2} - 2)}{(t^{2} - 2)(t-1)} \geqslant 0\\\\\\\dfrac{7t - 3 - 2t^{2}}{(t^{2} - 2)(t-1)} \geqslant 0$

ОДЗ: $\left \{ {\bigg{t^{2} - 2 \neq 0} \atop \bigg{t - 1 \neq 0 \ }} \right. \ \ \ \ \ \ \ \ \ \ \ \left \{ {\bigg{t \neq \pm \sqrt{2} } \atop \bigg{t \neq 1 \ \ \ \ }} \right.$

$7t - 3 - 2t^{2} = 0\\2t^{2} - 7t + 3 = 0\\D = (-7)^{2} - 4 \cdot 2 \cdot 3 = 49 - 24 = 25\\t_{1} = \dfrac{1}{2}\\\\t_{2} = 3\\$

По методу интервалов выясняем знаки неравенства и получаем:

$\left[\begin{array}{ccc}t < -\sqrt{2}\\\left \{ {\bigg{t \geqslant \dfrac{1}{2} } \atop \bigg{t < 1}} \right. \\\left \{ {\bigg{t \sqrt{2}} \atop \bigg{t \leqslant 3}} \right.\end{array}\right$

Обратная замена:

$\left[\begin{array}{ccc}3^{x} < -\sqrt{2}\\\left \{ {\bigg{3^{x} \geqslant \dfrac{1}{2} } \atop \bigg{3^{x} < 1}} \right. \\\left \{ {\bigg{3^{x} \sqrt{2}} \atop \bigg{3^{x} \leqslant 3}} \right.\end{array}\right \ \ \ \ \ \ \ \ \ \ \ \ \ \left[\begin{array}{ccc}x \in \O \ \ \ \ \ \ \ \ \ \\\left \{ {\bigg{x \geqslant \log_{3}\dfrac{1}{2} } \atop \bigg{x < 0 \ \ \ \ \ \ }} \right. \\\left \{ {\bigg{x \log_{3}\sqrt{2}} \atop \bigg{x \leqslant 1 \ \ \ \ \ \ \ }} \right.\end{array}\right$

$\left[\begin{array}{ccc}x \in \O \ \ \ \ \ \ \ \ \ \ \ \ \ \\x \in \bigg[\log_{3}\dfrac{1}{2}; \ 0 \bigg) \\x \in \(\log_{3}\sqrt{2}; \ 1] \ \ \end{array}\right$