Răspuns:
Notam [tex]A=\sqrt{26}+\sqrt{6}, B=\sqrt{13}+\sqrt{17}[/tex].
Atunci [tex]A^2 = 26+2\sqrt{26}\sqrt{6}+6 = 32+4\sqrt{39}[/tex]
[tex] B^2=13+2\sqrt{13}\sqrt{17}+17=30+2\sqrt{221}.[/tex]
Fie [tex] C = (A^2-30)/2 = 1 + 2\sqrt{39}, D=(B^2-30)/2=\sqrt{221} [/tex]
Pentru a compara A si B este suficient sa comparam C si D.
Avem [tex] C^2= 1 + 4\cdot 39 + 3\sqrt{39} = 157 + 3\sqrt{39}, D^2=221 [/tex]
Observam ca [tex] C^2 = 157 +3\sqrt{39}<157+2\sqrt{49}=156+14=170<221=D^2[/tex]
In concluzie, [tex]C^2<D^2[/tex], deci [tex]C<D[/tex], deci [tex]A^2-30<B^2-30[/tex], deci A<B.
