Question

Aratati ca $\frac{1}{1+ \sqrt{2} } + \frac{1}{ \sqrt{2}+ \sqrt{3} } + \frac{1}{ \sqrt{3}+2 }$ este numar natural.

Answer 1

Aplifici  fractiile  cu  conjugatul  numitorului
(1-√2)/(1+√2)*(1-√2)+(√2-√3)/(√2_√3)*(V2+V3)+(V3-V2)/(V3-V2)*(V3+V2)=
1-V2)/(1-2)+(V2-V3)/(2-3)+(V3-2)/(3-4)=
V2-1+V3-V2-(V3-2)=1∈N

Answer 2

$\sf ^{1-\sqrt2)}\frac{1}{1+ \sqrt{2} } +^{\sqrt2-\sqrt3)} \frac{1}{ \sqrt{2}+\sqrt{3} } + ^{\sqrt3-2)}\frac{1}{ \sqrt{3}+2 }= \\ \\ \frac{1-\sqrt2}{(1-\sqrt2)(1+\sqrt2)}+\frac{\sqrt2-\sqrt3}{(\sqrt2-\sqrt3)(\sqrt2+\sqrt3)}+\frac{\sqrt3-2}{(\sqrt3-2)(\sqrt3+2)}= \\ \\ \frac{1-\sqrt2}{1-2}+\frac{\sqrt2-\sqrt3}{2-3}+\frac{\sqrt3-2}{3-4}= \\ \\ -1+\sqrt2-\sqrt2+\sqrt3-\sqrt3+2=1 \\ \\ 1 \in{N$

Cu drag, Amalia Georgiana Monica Andreea Florica de la Cluj!