[tex]\frac{\sqrt{6}*\sqrt{24}}{\sqrt{4}}-(\frac{\sqrt{3}+0,5\sqrt{3}}{9})^2=\frac{\sqrt{6*24} }{2}-(\frac{\not{1,5}^{0,5}\sqrt{3}}{\not{9}^3})^2 =\frac{\sqrt{144} }{2}-(\frac{ 0,5\sqrt{3}}{3})^2 =\\\\=\frac{12}{2}-\frac{0,25*\not{3}^1}{\not{9}^3}=6-\frac{25}{100}:3=6-\frac{1}{4}*\frac{1}{3}=6-\frac{1}{12}=5\frac{11}{12}[/tex]
odp. Jest to liczba wymierna.
