Question

*Aratati ca numarul n=1/√3-1-1/√3+1,este natural.
Vreau rezolvarea completa va rog.

Answer 1

   
[tex]\displaystyle\\ n=\frac{1}{\sqrt{3}-1}-\frac{1}{\sqrt{3}+1}=~~~\text{(Rationalizam numitorii.)}\\\\ =\frac{\sqrt{3}+1}{(\sqrt{3}-1)(\sqrt{3}+1)}-\frac{\sqrt{3}-1}{(\sqrt{3}+1)(\sqrt{3}-1)}=\\\\ = \frac{\sqrt{3}+1}{\Big(\sqrt{3}\Big)^2-1^2}-\frac{\sqrt{3}-1} {\Big(\sqrt{3}\Big)^2-1^2}=\\\\ =\frac{\sqrt{3}+1}{3-1}-\frac{\sqrt{3}-1} {3-1}=\\\\ =\frac{\sqrt{3}+1}{2}-\frac{\sqrt{3}-1} {2}=\\\\ =\frac{\sqrt{3}+1-(\sqrt{3}-1)}{2}=\frac{\sqrt{3}+1-\sqrt{3}+1}{2}= \frac{1+1}{2}= \frac{2}{2}=\boxed{\bf 1\in N} [/tex]