Выражения под корнями взаимно обратные. Сделаем замену :
0" alt="(\sqrt{3-2\sqrt{2} })^{x}=m,m>0" align="absmiddle" class="latex-formula">
Тогда :
+ - +
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)](https://tex.z-dn.net/?f=1%29%28%5Csqrt%7B3-2%5Csqrt%7B2%7D%20%7D%29%5E%7Bx%7D%20%5Cleq3-2%5Csqrt%7B2%7D%5C%5C%5C%5C%283-2%5Csqrt%7B2%7D%29%5E%7B%5Cfrac%7Bx%7D%7B2%7D%20%7D%5Cleq%203-2%5Csqrt%7B2%7D%5C%5C%5C%5C%5Cfrac%7Bx%7D%7B2%7D%20%5Cleq1%5C%5C%5C%5Cx%5Cleq2%5C%5C%5C%5Cx%5Cin%28-%5Cinfty%3B2%5D%5C%5C%5C%5C2%29%28%5Csqrt%7B3-2%5Csqrt%7B2%7D%20%7D%29%5E%7Bx%7D%5Cgeq3%2B2%5Csqrt%7B2%7D%5C%5C%5C%5C%283-2%5Csqrt%7B2%7D%29%5E%7B%5Cfrac%7Bx%7D%7B2%7D%20%7D%5Cgeq%20%283-2%5Csqrt%7B2%7D%29%5E%7B-1%7D%5C%5C%5C%5C%5Cfrac%7Bx%7D%7B2%7D%5Cgeq-1%5C%5C%5C%5Cx%5Cgeq%20-2%5C%5C%5C%5Cx%5Cin%5B-2%3B%2B%5Cinfty%29 "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]