Question

ДОПОМОЖІТЬ ОЧЕНЬ СРОЧНО!!!!!!!!!!!

Answer 1

Ответ:

$\left ( - \infty ~ ; ~- \log_{5} \dfrac{63 + 4\sqrt{209} }{125} \right ] \bigcup \left[-\log_5\dfrac{63 - 4\sqrt{209} }{125} ~ ; ~ \infty \right )$

Объяснение:

$\displaystyle (0,2)^{2x-3} - 126 \cdot (0,2)^x + 5\geqslant 0 \\\\ 125\cdot \bigg( \frac{1}{5}\bigg) ^{2x} - 126 \cdot \bigg(\frac{1}{5} \bigg )^x +5 \geqslant 0$

Введем замену

$\displaystyle u=\bigg(\frac{1}{5}\bigg)^x ~ ; ~ u^2=\bigg(\frac{1}{5}\bigg)^{2x}$

$\displaystyle 125 u^2- 126 u +5 \geqslant 0 ~ \\\\\\ D = 126^2 -2500 = 13376$

$u_{1,2} = \dfrac{126 \pm 8\sqrt{209} }{250} = \dfrac{63 \pm 4\sqrt{209} }{125}$

$u_1 \approx 0,9 \\\\ u_2 \approx 0,04$

Отобразим на интервале

$\setlength{\unitlength}{23mm}\begin{picture}(1,1) \linethickness{0.2mm} \put(0.8,-0.3) {\sf 0,04} \put(.1 ,0.1){ \Large \hspace{-0,4em}\times\times\times \times } \put(2.1 ,0.1){ \Large \hspace{-0,4em}\times\times\times \times } \put(1,0){\circle*{0.05}} \put(1,0) {\line(0,2){0.3}} \put(1,0.3) {\vector(1,0){1.7}} \put(2,-0.3) {\sf 0,9}\put(2.05,0){\circle*{0.05}} \put(2.05,0) {\line(0,2){0.3}} \put(1,0.3) {\vector(-1,0){0.7} } \ \put(0,0){\vector (1,0){3}} \end{picture}$

Теперь подставим

$\left [\begin{array}{l}\bigg(\dfrac{1}{5}\bigg)^x \geqslant u_1 \\\\ \bigg(\dfrac{1}{5}\bigg)^x \leqslant u_2\end{array}\right. \\\\\\ \left [\begin{array}{l}\bigg(\dfrac{1}{5}\bigg)^x \geqslant \dfrac{63 + 4\sqrt{209} }{125} \\\\ \bigg(\dfrac{1}{5}\bigg)^x \leqslant \dfrac{63 - 4\sqrt{209} }{125}\end{array}$

$\left [\begin{array}{l} 5^{-x} \geqslant \dfrac{63 + 4\sqrt{209} }{125} \\\\ 5^{-x} \leqslant \dfrac{63 - 4\sqrt{209} }{125}\end{array}$

$\left [\begin{array}{l} {-x} \geqslant \log_{5} \dfrac{63 + 4\sqrt{209} }{125} \\\\ -x \leqslant \log_5\dfrac{63 - 4\sqrt{209} }{125}\end{array}$

$\left [\begin{array}{l} {x} \leqslant - \log_{5} \dfrac{63 + 4\sqrt{209} }{125} \\\\ x \geqslant- \log_5\dfrac{63 - 4\sqrt{209} }{125}\end{array}$