Răspuns:
a) [tex]\sqrt{5+4} + \sqrt{2*5+6} = \sqrt{9} +\sqrt{16} = 3+4 \\[/tex] = 7
b)
[tex]\sqrt{n+4}< 2\sqrt{7} \leq \sqrt{2n+6}[/tex] (Facem inegalitatile pe rand)
1.
[tex]\sqrt{n+4}<2\sqrt{7} \\ (\sqrt{n+4})^2<(2\sqrt{7})^2\\n+4 < 28\\n < 24[/tex]
2.
[tex]2\sqrt{7} \leq \sqrt{2n+6} \\(2\sqrt{7})^2 \leq (\sqrt{2n+6})^2 \\28 \leq 2n+6\\22\leq 2n\\11\leq n[/tex]
=> n apartine 11, 12, 13, 14,...,23
Explicație pas cu pas:
