Question

3-2\sqrt{2} } )^x+(\sqrt{3+2\sqrt{2} } )^x\geq 6[/tex] решите

Answer 1

Выражения под корнями взаимно обратные. Сделаем замену :

$(\sqrt{3-2\sqrt{2} })^{x}=m,m0$

Тогда :

$(\sqrt{3+2\sqrt{2} })^{x}=\frac{1}{m}$

$m+\frac{1}{m} \geq 6\\\\m^{2}-6m+1\geq0\\\\m^{2}-6m+1=0\\\\D=36-4=32=4\sqrt{2}\\\\m_{1}=\frac{6-4\sqrt{2} }{2}=3-2\sqrt{2}\\\\m_{2}=\frac{6+4\sqrt{2} }{2}=3+2\sqrt{2}$

$(m-(3-\sqrt{2}))(m-(3+2\sqrt{2}))\geq 0$

          +                                -                                   +

0_________[3-2√2]__________[3 + 2√2]__________  m

1) 0 < m ≤ 3 - 2√2                    2) m ≥ 3 + 2√2

$1)(\sqrt{3-2\sqrt{2} })^{x} \leq3-2\sqrt{2}\\\\(3-2\sqrt{2})^{\frac{x}{2} }\leq 3-2\sqrt{2}\\\\\frac{x}{2} \leq1\\\\x\leq2\\\\x\in(-\infty;2]\\\\2)(\sqrt{3-2\sqrt{2} })^{x}\geq3+2\sqrt{2}\\\\(3-2\sqrt{2})^{\frac{x}{2} }\geq (3-2\sqrt{2})^{-1}\\\\\frac{x}{2}\geq-1\\\\x\geq -2\\\\x\in[-2;+\infty)$

ответ : x ∈ [- 2 ; 2]