Cześć!
Rozwiązanie
[tex]3\sqrt3=\sqrt{3^2\cdot3}=\sqrt{9\cdot3}=\sqrt{27}\\\\2\sqrt6=\sqrt{2^2\cdot6}=\sqrt{4\cdot6}=\sqrt{24}\\\\\sqrt{29}\\\\4\sqrt2=\sqrt{4^2\cdot2}=\sqrt{16\cdot2}=\sqrt{32}\\\\\sqrt{24}<\sqrt{27}<\sqrt{29}<\sqrt{32}\\\\\huge\boxed{2\sqrt6<3\sqrt3<\sqrt{29}<\sqrt{32}}[/tex]
