DISEQUAZIONI NUMERICHE MISTE


${\begin{matrix} \sqrt{2 x^{2} - 3 x + 4} + 1 > 2 x \\ ∣ x - 1 ∣ - ∣ \frac{1}{2} x - 8 ∣ < x \end{matrix}$
${\begin{matrix} \frac{\sqrt{1 - x}}{2} > \frac{\sqrt{1 + x}}{3} \\ ∣ \frac{2 + 5 x}{1 - 2 x} ∣ + 2 > 7 \end{matrix}$
${\begin{matrix} ∣ \frac{x^{2}}{x - 1} ∣ + x > 3 \\ \sqrt{x^{2} - 5 x + 6} > x - 1 \end{matrix}$
${\begin{matrix} \frac{\sqrt{6 x - x^{2}}}{3 - 2 x} \geq 1 \\ ∣ \frac{x + 3}{x - 2} ∣ < 4 \end{matrix}$
${\begin{matrix} \sqrt{x} + \sqrt{x + 4} > \sqrt{2 x - 1} \\ ∣ - 2 x + 10 ∣ - 6 < 3 x + 3 (1 - x) \end{matrix}$
${\begin{matrix} \sqrt{x^{2} - 5 x + 7} \leq 2 x - 4 \\ ∣ x^{2} - 5 x + 6 ∣ \leq ∣ x - 2 ∣ \end{matrix}$
${\begin{matrix} \frac{∣ 4 x - 1 ∣ - 3 x - 1}{x^{2} + 4 x + 16} > 0 \\ \sqrt{x^{2} + 5 x - 14} > \sqrt{x^{2} + 4 x + 3} \end{matrix}$
${\begin{matrix} \sqrt{x + 2} < ∣ x ∣ \\ ∣ x + 1 ∣ + ∣ x ∣ \geq - 2 \end{matrix}$
${\begin{matrix} \frac{4 - \sqrt{x^{2} + 7 x + 10}}{x} < - 2 \\ x^{4} - 16 x^{2} \leq 0 \end{matrix}$
${\begin{matrix} \frac{1 - 2 x - \sqrt{2 - x}}{x^{2} + 5 x + 15} \geq 0 \\ \frac{3 - ∣ 4 x + 3 ∣}{x + 7 + \sqrt{x + 2}} < 0 \end{matrix}$
${\begin{matrix} \frac{x^{4} - 6 x^{2} + 9}{x - 1} \leq 0 \\ \frac{∣ x - 8 ∣ - 1}{\sqrt{x + 2}} > 0 \end{matrix}$
${\begin{matrix} \frac{3 x - \sqrt{x^{2} - 1}}{∣ x ∣ + \sqrt{x + 5}} \geq 0 \\ \sqrt{- 3 x^{2} + 2 x + 1} \geq 0 \end{matrix}$
${\begin{matrix} \frac{x^{2} - 4 x + 4}{x - \sqrt{x^{2} - 1}} > 0 \\ ∣ 2 x - 1 ∣ - ∣ x + 3 ∣ \leq 2 - 3 x \end{matrix}$
${\begin{matrix} \frac{∣ x^{2} - 3 x ∣ - 10}{∣ x ∣} \leq 3 \\ \frac{\sqrt[3]{x^{3} + 1} - \sqrt{x^{2} + 1}}{x - 2} \geq 0 \end{matrix}$