[tex]\displaystyle{\frac{7}{17} < \frac{17}{n^{2} } < \frac{13}{17} }[/tex]
[tex]\displaystyle{\frac{7*17}{17} < \frac{17*17}{n^{2} } < \frac{13*17}{17} }[/tex]
[tex]\displaystyle{7 < \frac{289}{n^{2} }< 13 }[/tex]
[tex]\displaystyle {7^{-1} > (\frac{289}{n^{2} })^{-1} > 13^{-1}}[/tex]
[tex]\displaystyle{\frac{1}{7}>\frac{n^{2} }{289}>\frac{1}{13} }[/tex]
[tex]\displaystyle{\frac{1}{13} < \frac{n^{2}}{289} < \frac{1}{7}}[/tex]
[tex]\displaystyle{\frac{1*7*289}{13*7*289}< \frac{13*7*n^{2} }{13*7*289}< \frac{1*13*289}{7*13*289} }[/tex]
[tex]\displaystyle {\frac{2023}{26299}< \frac{91n^{2} }{26299}<\frac{3757}{26299} }[/tex]
[tex]\displaystyle {\frac{2023*26299}{26299}< \frac{91*26299n^{2} }{26299}<\frac{3757*26299}{26299} }[/tex]
[tex]\displaystyle{2023<91n^{2} < 3757}[/tex]
[tex]\displaystyle{2023:91<91:91n^{2} < 3757:91}[/tex]
[tex]\displaystyle{22,23 < n^{2} <41,28}[/tex]
[tex]\displaystyle{\sqrt{22,23}< \sqrt{n^{2} }< \sqrt{41,28} }[/tex]
4,71 < [tex]\displaystyle {n}[/tex] < 6,42
[tex]\displaystyle {n}[/tex] aparține mulțimii numerelor naturale.
Mulțimea numerelor naturale este N = {0, 1, 2, 3, 4, 5, ......, etc, .....}
Prin urmare, în mulțimea N, numărul [tex]\displaystyle {n}[/tex] va lua valorile 5 și 6.
Soluție: [tex]\displaystyle {n}[/tex] ∈ {5, 6}
- Lumberjack25
