[tex]a) \ \sqrt{\dfrac{49 + 49^2}{50 + 50^2} } = \sqrt{\dfrac{49 \cdot (1 + 49)}{50 \cdot (1 + 50)} } = \sqrt{\dfrac{49 \cdot 50}{50 \cdot 51} } = \sqrt{\dfrac{49}{51} }[/tex]
[tex]\sqrt{\dfrac{50 + 50^2}{51 + 51^2} } = \sqrt{\dfrac{50 \cdot (1 + 50)}{51 \cdot (1 + 51)} } = \sqrt{\dfrac{50 \cdot 51}{51 \cdot 52} } = \sqrt{\dfrac{50}{52} }[/tex]
Deoarece
[tex]49 \cdot 52 < 51 \cdot 50\Leftrightarrow 2548 < 2550[/tex]
[tex]\Rightarrow \dfrac{49}{51} < \dfrac{50}{52} \Rightarrow \sqrt{\dfrac{49}{51}} < \sqrt{\dfrac{50}{52}}[/tex]
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[tex]b) \ \sqrt{ \dfrac{16 + \sqrt{192}} {18 + \sqrt{243} } } = \sqrt{ \dfrac{16 + 8\sqrt{3}} {18 + 9\sqrt{3} } } = \sqrt{ \dfrac{8(2+\sqrt{3})} {9(2+\sqrt{3}) } } = \sqrt{ \dfrac{8}{9} }[/tex]
[tex]\sqrt{ \dfrac{7 + \sqrt{147}} {8 + \sqrt{192} } } = \sqrt{ \dfrac{7 + 7\sqrt{3}} {8 + 8\sqrt{3} } } = \sqrt{ \dfrac{7(1+\sqrt{3})}{8(1+\sqrt{3}) } } = \sqrt{ \dfrac{7}{8} }[/tex]
Deoarece
[tex]8 \cdot 8 > 9 \cdot 7 \Leftrightarrow 64 > 63[/tex]
[tex]\Rightarrow \dfrac{8}{9} > \dfrac{7}{8} \Rightarrow \sqrt{\dfrac{8}{9}} > \sqrt{\dfrac{7}{8}}[/tex]
q.e.d.
