[tex]\log_{1/2}\frac{3x-1}{x^2+1} > 0\Leftrightarrow \frac{3x-1}{x^2+1} < 1\Leftrightarrow 3x-1-x^2-1 < 0\Leftrightarrow -x^2+3x-2 < 0\Leftrightarrow \\\Leftrightarrow (x-2)(1-x) < 0\Rightarrow x\in \left ( -\infty ,1 \right )\cup \left ( 2,\infty \right )\\\frac{3x-1}{x^2+1} > 0\Leftrightarrow 3x-1 > 0\Rightarrow x > \frac{1}{3}[/tex][tex]\left \{ x\in \left ( -\infty ,1 \right )\cup \left ( 2,\infty \right ) \right \}\cap \left \{ x > \frac{1}{3} \right \}=\left \{ x\in\left ( \frac{1}{3},1 \right )\cup \left ( 2,\infty \right ) \right \}[/tex]
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[tex]\log_{1/2}\frac{3x-1}{x^2+1} > 0\Leftrightarrow \frac{3x-1}{x^2+1} < 1\Leftrightarrow 3x-1-x^2-1 < 0\Leftrightarrow -x^2+3x-2 < 0\Leftrightarrow \\\Leftrightarrow (x-2)(1-x) < 0\Rightarrow x\in \left ( -\infty ,1 \right )\cup \left ( 2,\infty \right )\\\frac{3x-1}{x^2+1} > 0\Leftrightarrow 3x-1 > 0\Rightarrow x > \frac{1}{3}[/tex][tex]\left \{ x\in \left ( -\infty ,1 \right )\cup \left ( 2,\infty \right ) \right \}\cap \left \{ x > \frac{1}{3} \right \}=\left \{ x\in\left ( \frac{1}{3},1 \right )\cup \left ( 2,\infty \right ) \right \}[/tex]