2) [tex]x + \frac{1}{x} = 3[/tex]
Ridicam la patrat aceasta egalitate:
[tex](x+\frac{1}{x})^2=3^2\ \textless \ =\ \textgreater \ x^2+2*x*\frac{1}{x}+(\frac{1}{x})^2=9 \ \textless \ =\ \textgreater \ \\ \ \textless \ =\ \textgreater \ x^2+2+\frac{1}{x^2}=9\ \textless \ =\ \textgreater \ x^2+\frac{1}{x^2}=9-2\ \textless \ =\ \textgreater \ x^2+\frac{1}{x^2}=7[/tex]
3) Aducem fiecare termen al sumei la o forma mai simpla:
[tex]\sqrt{5}-3\ \textless \ 0 \ \textless \ =\ \textgreater \ \sqrt{5}\ \textless \ 3\ \textless \ =\ \textgreater \ \sqrt{5}\ \textless \ \sqrt{9}(Adevarata) =\ \textgreater \ \\ =\ \textgreater \ |\sqrt{5}-3|=3-\sqrt{5}[/tex]
Rationalizam numitorul:
[tex]\frac{4}{3-\sqrt{5}}=\frac{4(3+\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})} = \frac{4(3+\sqrt{5})}{4}=3+\sqrt{5}[/tex]
Atunci:
[tex]a = 3-\sqrt{5}+3+\sqrt{5}=\ \textgreater \ a=6\in \mathbb{N}[/tex]
Succes
