Question

aratati ca:
$\frac{1}{1+ \sqrt{2} } + \frac{1}{ \sqrt{2}+ \sqrt{3}} + \frac{1}{ \sqrt{3} +2}$
⊂N.

Answer 1

$\frac{1}{1+ \sqrt{2} } + \frac{1}{ \sqrt{2} + \sqrt{3} } + \frac{1}{ \sqrt{3} +2} =$

Amplific fiecare fractie cu conjugata sa, adica cu:$1- \sqrt{2} ; \sqrt{2} - \sqrt{3} ; \sqrt{3} -2$ si obtin:

[tex] \frac{1 -\sqrt{2} }{1-2} + \frac{ \sqrt{2} - \sqrt{3} }{2-3} + \frac{ \sqrt{3} -2}{3-4} = =-1(1- \sqrt{2})-1( \sqrt{2} - \sqrt{3} ) -1( \sqrt{3} - 2 )= =-1 + \sqrt{2} - \sqrt{2} + \sqrt{3} - \sqrt{3} +2= =1 [/tex] ∈ N

Answer 2

[tex]\frac{1}{1+\sqrt2}+\frac{1}{\sqrt2+\sqrt3}+\frac{1}{\sqrt3+2}\;=\;...\;rationalizam\;!\\ =\frac{1-\sqrt2}{-1}+\frac{\sqrt2-\sqrt3}{-1}+\frac{\sqrt3-2}{-1}=\frac{1-2}{-1}=(-1)/(-1)=1[/tex]