Question

решите без логарифмов
${( \sqrt{2} - 1) }^{x} + {( \sqrt{2} + 1)}^{x} \geqslant 6$
​

Answer 1

Ответ:

$(\sqrt2-1)^{x}+(\sqrt2+1)^{x}\geq 6\\\\\sqrt2-1=\dfrac{(\sqrt2-1)(\sqrt2+1)}{\sqrt2+1}=\dfrac{2-1}{\sqrt2+1}=\dfrac{1}{\sqrt2+1}\\\\\\t=(\sqrt2+1)^{x}>0\ \ ,\ \ \ t+\dfrac{1}{t} \geq 6\ \ ,\ \ \ t+\dfrac{1}{t}-6\geq 0\ \ ,\ \ \dfrac{t^2-6t+1}{t}\geq 0\ ,\\\\\\t^2-6t+1=0\ \ ,\ \ D/4=8\ ,\\\\t_1=3-\sqrt8=3-2\sqrt2\approx 0,17\ \ ,\ \ t_2=3+2\sqrt2\approx 5,83\\\\\dfrac{(t-3+2\sqrt2)(t-3-2\sqrt2)}{t}\geq 0\ ,\\\\znaki:\ \ ---(0)+++[\, 3-2\sqrt2\, ]---[\, 3+2\sqrt2\, ]+++\ \ \ \ t>0\ ,$

$t\in (\ 0\ ;\ 3-2\sqrt2\ ]\cup [\ 3+2\sqrt2\ ;+\infty )\ \ \to \ \ \left[\begin{array}{l}(0<(\sqrt2+1)^{x}\leq 3-2\sqrt2\\(\sqrt2+1)^{x}\geq 3+2\sqrt2\end{array}\right\\\\\\\left[\begin{array}{l}x\leq log_{\sqrt2+1}(3-2\sqrt2)\\x\geq log_{\sqrt2+1}(3+2\sqrt2)\end{array}\right\\\\\\Otvet:\ \ x\in (-\infty \ ;\, log_{\sqrt2+1}(3-2\sqrt2)\ ]\cup [\ log_{\sqrt2+1}(3+2\sqrt2)\ ;+\infty \ )\ .$