Question

Aratati ca numarul 1/1+√2+ 1/√2+√3+ 1/√3+2 este natural. Va roog!!

Answer 1

   
[tex]\displaystyle \\ \frac{1}{1+ \sqrt{2}} +\frac{1}{ \sqrt{2}+\sqrt{3}} +\frac{1}{\sqrt{3}+2} = \\ \\ ~~~ =\frac{1}{\sqrt{2}+1} +\frac{1}{ \sqrt{3}+\sqrt{2}} +\frac{1}{2+\sqrt{3}} = \text {(Rationalizam numitorul)}\\ \\ =\frac{1(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)} +\frac{1(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})} +\frac{1(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})} = [/tex]


[tex]\displaystyle \\ =\frac{\sqrt{2}-1}{(\sqrt{2})^2-1^2} +\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3})^2-(\sqrt{2})^2} +\frac{2-\sqrt{3}}{2^2-(\sqrt{3})^2} = \\ \\ =\frac{\sqrt{2}-1}{2-1} +\frac{\sqrt{3}-\sqrt{2}}{3-2} +\frac{2-\sqrt{3}}{4-3} = \\ \\ =\frac{\sqrt{2}-1}{1} +\frac{\sqrt{3}-\sqrt{2}}{1} +\frac{2-\sqrt{3}}{1} = \\ \\ =\sqrt{2}-1+\sqrt{3}-\sqrt{2}+2-\sqrt{3}= \\ \\ =\sqrt{3}-\sqrt{3}+\sqrt{2}-\sqrt{2}+2-1 = 2-1 = \boxed{1 \in N}[/tex]