Question

помогите решить уравнение ​

Answer 1

Ответ:

Объяснение:

$4 * (\frac{1}{16})^x - 17 * (\frac{1}{4})^x + 4 = 0\\(\frac{1}{4})^x = t > 0\\4t^2 - 17t + 4 = 0\\t_1 = 4\\t_2 = \frac{1}{4}\\(\frac{1}{4})^x = t_1 = 4\\x_1 = -1.\\(\frac{1}{4})^x = t_2 = \frac{1}{4}\\x_2 = 1.\\Answer: \\1\\-1$

Answer 2

$\displaystyle 4\cdot (\frac{1}{16})^x-17\cdot(\frac{1}{4})^x +4=0\\\\ 4\cdot (\frac{1}{4})^{2x}-17\cdot (\frac{1}{4})^x+4=0 \\\\ (\frac{1}{4})^x=y; \;\;\;\; y>0\\\\ 4y^2-17y+4=0\\ D=(17)^2-4\cdot 4\cdot 4=289-64=(15)^2;$

$\displaystyle D>0 \Rightarrow$ 2 р.д.к.

$\displaystyle y_{1,2}=\frac{17б15}{8}=\left |{ {{\frac{1}{4}} \atop {4}} \right. \\\\ \left [{ {{y=\frac{1}{4}} \atop {y=4}} \right. \left [{ {{(\frac{1}{4})^x=\frac{1}{4}} \atop {(\frac{1}{4})^x=4}} \right. \left [{ {{x=1} \atop {x=-1}} \right.$

Ответ:  $-1;1$