N° 10 SISTEMI DISEQUAZIONI

${\begin{matrix} \frac{1 - 2 x - \sqrt{2 - x}}{x^{2} + 5 x + 15} \geq 0 \\ \frac{3 - ∣ 4 x + 3 ∣}{x + 7 + \sqrt{x + 2}} < 0 \end{matrix}$

1 − 2 x − 2 − x x 2 + 5 x + 15 ≥ 0
- $1 - 2 x - \sqrt{2 - x} \geq 0 \Rightarrow \sqrt{2 - x} \leq 1 - 2 x \Rightarrow {\begin{matrix} 1 - 2 x \geq 0 \\ 2 - x \geq 0 \\ 2 - x \leq {(1 - 2 x)}^{2} \end{matrix} \Rightarrow {\begin{matrix} x \leq \frac{1}{2} \\ x \leq 2 \\ 2 - x \leq 1 + 4 x^{2} - 4 x \end{matrix} \Rightarrow {\begin{matrix} x \leq \frac{1}{2} \\ x \leq 2 \\ 4 x^{2} - 3 x - 1 \geq 0 \end{matrix} \Rightarrow$
  $\Rightarrow {\begin{matrix} x \leq \frac{1}{2} \\ x \leq 2 \\ x \leq - \frac{1}{4} \lor x \geq 1 \end{matrix} \Rightarrow$ $x \leq - \frac{1}{4}$
- $x^{2} + 5 x + 15 > 0 \Rightarrow$ $\forall x \in ℝ$
$x \leq - \frac{1}{4}$
3 − ∣ 4 x + 3 ∣ x + 7 + x + 2 < 0 ⇒ ∣ 4 x + 3 ∣ − 3 x + 7 + x + 2 > 0
- $∣ 4 x + 3 ∣ - 3 > 0 \Rightarrow {\begin{matrix} 4 x + 3 \geq 0 \\ 4 x + 3 - 3 > 0 \end{matrix} \lor {\begin{matrix} 4 x + 3 < 0 \\ - 4 x - 3 - 3 > 0 \end{matrix} \Rightarrow {\begin{matrix} x \geq - \frac{3}{4} \\ x > 0 \end{matrix} \lor {\begin{matrix} x < - \frac{3}{4} \\ x < - \frac{3}{2} \end{matrix} \Rightarrow$ $x < - \frac{3}{2} \lor x > 0$
- $x + 7 + \sqrt{x + 2} > 0 \Rightarrow \sqrt{x + 2} > - (x + 7) \Rightarrow {\begin{matrix} - x - 7 \geq 0 \\ x + 2 > {(x + 7)}^{2} \end{matrix} \lor {\begin{matrix} - x - 7 < 0 \\ x + 2 \geq 0 \end{matrix} \Rightarrow {\begin{matrix} x \leq - 7 \\ x + 2 > x^{2} + 49 + 14 x \end{matrix} \lor {\begin{matrix} x > - 7 \\ x \geq - 2 \end{matrix} \Rightarrow$
  $\Rightarrow {\begin{matrix} x \leq - 7 \\ x^{2} + 13 x + 47 < 0 \end{matrix} \lor x \geq - 2 \Rightarrow {\begin{matrix} x \leq - 7 \\ ∄ x \in ℝ \end{matrix} \lor x \geq - 2 \Rightarrow ∄ x \in ℝ \lor x \geq - 2 \Rightarrow$ $x \geq - 2$
$- 2 \leq x < - \frac{3}{2} \lor x > 0$

$- 2 \leq x < - \frac{3}{2}$