Question

De rezolvat exercițiile 2 și 3

Answer 1

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