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## Mathematics: Inequalities ### Solution

$\left(\frac{1}{3}\right)^{\frac{|x|+1}{x-2}}>9$ $\left(\frac{1}{3}\right)^{\frac{|x|+1}{x-2}}>\left(\frac{1}{3}\right)^{-2}$ The expression exists when, $\frac{|x|+1}{x-2}<-2$ Now, we focus on the absolute value of $$x$$  If $$x<0$$ $\frac{-x+1+2 x-4}{x-2}<0$ $\frac{x-3}{x-2}<0$ $\frac{(x-3)(x-2)}{(x-2)^{2}}<0$ $\Rightarrow 2<x<3$ We have a contradiction for this case.  If $$x\geq0$$ $\frac{x+1+2 x-4}{x-2}<0$ $\frac{3x-3}{x-2}<0$ $\frac{(3x-3)(x-2)}{(x-2)^{2}}<0$ $\frac{3(x-1)(x-2)}{(x-2)^{2}}<0$ $\Rightarrow 1<x<2$ For this case we have solutions.  Finally, $\huge S_{\mathbb{R}}=[0,+\infty[\cap] 1,2[=] 1,2[$
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