Zad. 9
[tex]A = \{x \in R: x \leq \sqrt{13-4\sqrt{3}} - \sqrt{21-12\sqrt{3}} \} \\\\ x \leq \sqrt{13-4\sqrt{3}} - \sqrt{21-12\sqrt{3}} = \sqrt{(2\sqrt{3}-1)^2} - \sqrt{(2\sqrt{3}-3)^2} = \\ = 2\sqrt{3}-1 - (2\sqrt{3}-3)=2\sqrt{3}-1 -2\sqrt{3}+3 = 2 \\ x \in (- \infty, \ 2 \rangle \\\\ Zatem: \\ A = \{x \in R: x \in (- \infty, \ 2 \rangle \}[/tex]
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[tex]B = \{x \in R: |2x + 12| \geq 2 \} \\\\ |2x + 12| \geq 2 \\ 2x + 12 \geq 2 \ \vee \ 2x + 12 \leq -2 \\ 2x + 12 \geq 2 \ \vee \ 2x + 12 \leq -2 \\ 2x \geq 2-12 \ \vee \ 2x \leq -2-12 \\ 2x \geq -10 \ \ \ |:2 \ \vee \ 2x \leq -14 \\ x \geq -5 \ \vee \ x \leq - 7 \\ x \in (- \infty, \ - 7 \rangle \cup \langle -5, \ + \infty) \\\\ Zatem: \\ B = \{x \in R: x \in (- \infty, \ - 7 \rangle \cup \langle -5, \ + \infty) \}[/tex]
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[tex]B \setminus A =((- \infty, \ - 7 \rangle \cup \langle -5, \ + \infty)) \setminus (- \infty, \ 2 \rangle = (2, \ + \infty) \\\\ Zatem: \\ B \setminus A = \{x \in R: x \in (2, \ + \infty) \}[/tex]
