$\{\begin{array}{c}\frac{x^2-4x+4}{x-\sqrt{x^2-1}}>0\\ \mid 2x-1\mid -\mid x+3\mid \le 2-3x\end{array}$

• $\frac{x^2-4x+4}{x-\sqrt{x^2-1}}>0$
  • $x^2-4x+4>0\Rightarrow {\left(x-2\right)}^2>0\Rightarrow x\ne 0$
  • $x-\sqrt{x^2-1}>0\Rightarrow \sqrt{x^2-1}<x\Rightarrow \{\begin{array}{c}x\ge 0\\ x^2-1\ge 0\\ x^2-1<x^2\end{array}\Rightarrow \{\begin{array}{c}x\ge 0\\ x\le -1\vee x\ge 1\\ \forall x\in ℝ\end{array}\Rightarrow x\ge 1$

    $x\ge 1$

• $\mid 2x-1\mid -\mid x+3\mid \le 2-3x\Rightarrow \{\begin{array}{c}x\le -3\\ -2x+1+x+3\le 2-3x\end{array}\vee \{\begin{array}{c}-3\le x\le \frac{1}{2}\\ -2x+1-x-3\le 2-3x\end{array}\vee \{\begin{array}{c}x\ge \frac{1}{2}\\ 2x-1-x-3\le 2-3x\end{array}\Rightarrow$

  $\Rightarrow \{\begin{array}{c}x\le -3\\ x\le \frac{1}{2}\end{array}\vee \{\begin{array}{c}-3\le x\le \frac{1}{2}\\ \forall x\in ℝ\end{array}\vee \{\begin{array}{c}x\ge \frac{1}{2}\\ x\le \frac{3}{2}\end{array}\Rightarrow x\le -3\vee -3\le x\le \frac{1}{2}\vee \frac{1}{2}\le x\le \frac{3}{2}\Rightarrow$$x\le \frac{3}{2}$

$1\le x\le \frac{3}{2}$