1<(3+√2)/(2+√3)<2 [latex]it 1 extless dfrac{3+sqrt2}{2+sqrt3} extless 2 Leftrightarrow 2 + sqrt3 extless 3+ sqrt2 extless 4+2sqrt3[/latex] Avem o dublă inegalitate, prin urmare vom avea două cazuri: [latex]it I) 2+sqrt3 extless 3+sqrt2 Leftrightarrow (2+sqrt3)^2 extless (3+sqrt2)^2 Leftrightarrow \;\ Leftrightarrow 5+ 4sqrt3 extless 5+6sqrt2 Leftrightarrow 4sqrt{3} extless 6sqrt{2} Lefrightarrow sqrt{4^2cdot3} extless sqrt{6^2cdot2} Leftrightarrow \;\ Leftrightarrow sqrt{48} extless sqrt{72} (A)[/latex]