[tex]\it Not\breve{a}m\ x=a,\ \ \dfrac{1}{x}=b,\ cu\ condi\c{\it t}ia\ evident\breve{a}\ ab=1\ \ \ \ (*).\\ \\ \\ Rela\c{\it t}ia\ dat\breve{a}\ devine:\\ \\ \\ a+b=6 \Rightarrow (a+b)^2=6^2\Rightarrow a^2+2ab+b^2=36 \stackrel{(*)}{\Longrightarrow}a^2+2+b^2=36|_{-2} \Rightarrow\\ \\ \\ \Rightarrow a^2+b^2=34|_{-2ab} \Rightarrow a^2+b^2-2ab=34-2ab \stackrel{(*)}{\Longrightarrow} (a-b)^2 =32 \Rightarrow[/tex]
[tex]\it \Rightarrow \sqrt{(a-b)^2}=\sqrt{32} \Rightarrow \sqrt{(a-b)^2}=\sqrt{16\cdot2} \Rightarrow \sqrt{(a-b)^2}=4\sqrt{2} \Rightarrow \\ \\ \\ \Rightarrow |a-b| = 4\sqrt2 \Rightarrow a-b=\pm4\sqrt2.[/tex]
Revenind asupra notației, rezultă:
[tex]\it x-\dfrac{1}{x}=\pm4\sqrt2[/tex]
