Question

буду благодарна за помощь!! ​

Answer 1

Ответ:

$\displaystyle(-\infty;8]$

Пошаговое объяснение:

$\displaystyle3(\sqrt 2)^x-7\cdot 2^{\frac{x}{4}}-20\le 0\\\\3(2^{\frac{1}{2}})^x-7\cdot (2^{\frac{x}{4}}-20\le 0\\\\3(2^{\frac{1}{2}}x)-7\cdot (2^{\frac{x}{4}}-20\le 0\\\\3(2^{2\cdot \frac{1}{4}x})-7\cdot (2^{\frac{x}{4}}-20\le 0\\\\3(2^{\frac{1}{4}x})^2-7\cdot (2^{\frac{x}{4}}-20\le 0$

замінюємо

$\displaystyle2^{\frac{1}{4}x}=t,\ t > 0\\\\3t^2-7t-20\le 0$

Нули:

$\displaystyle3t^{2} - 7t - 20 =0\\\\ a=3 ,\ \ b=-7 ,\ \ c=-20\\\\ D = b^2 - 4ac = ( - 7)^2 - 4\cdot3\cdot( - 20) = 49 + 240 = 289\\\\\sqrt{D} =\sqrt{289} = 17\\\\ t_1=\frac{-b-\sqrt{D}}{2a}=\frac{7-17}{2\cdot3}=\frac{-10 }{6 }=- \frac{ 5 }{ 3 }=-1\frac{2}{3} \\\\ t_2=\frac{-b+\sqrt{D}}{2a}=\frac{7+17}{2\cdot3}=\frac{24}{6}=4\\\\t\in \left[ -1 \frac{ 2 }{ 3 } ;4 \right]\\\\\underline{t\in \left(0 ;4 \right]}$

$\displaystyle2^{\frac{1}{4}x}\in\left(0;4\right]\\\\1)\\\\2^{\frac{1}{4}x} > 0\\\\x\in R\\\\2)\\\\2^{\frac{1}{4}x}\le 4\\\\2^{\frac{1}{4}x}\le 2^2\\\\\frac{1}{4}x\le 2\ \ \ \ |\cdot 4\\\\x\le 8$

Ответ:

$\displaystyle x\in(-\infty;8]$