Question

aratati ca daca x=$\sqrt{3}$ + 2 , atunci $x^{2}$ + $\frac{1}{ x^{2} }$ este numar natural

Answer 1

[tex] x^{2} + \frac{1}{ x^{2} } =( \sqrt{3}+2)^{2}+ \frac{1}{(\sqrt{3}+2)^{2} }= \ \textgreater \ \\ 3+2 \sqrt{3} +4+\frac{(\sqrt{3}-2)^{2}}{(\sqrt{3}+2)^{2}(\sqrt{3}-2)^{2}}=\ \textgreater \ \\ 7+2\sqrt{3} + \frac{3-2\sqrt{3}+4}{(3-4)^{2}} = 7+2\sqrt{3}+7-2\sqrt{3}=14[/tex]