Odpowiedź:
[tex]Zad.23\\\\P=a^2\\\\P=(9-3\sqrt{5})^2=9^2-2\cdot9\cdot3\sqrt{5}+(3\sqrt{5})^2=81-54\sqrt{5}+9\cdot5=81-54\sqrt{5}+45=\\\\=126-54\sqrt{5}\\\\Odp.D\\\\\\Zad.24\\\\2\sqrt{3}-\dfrac{\sqrt{3}}{2\sqrt{3}-2}=2\sqrt{3}-\dfrac{\sqrt{3}}{2\sqrt{3}-2}\cdot\dfrac{2\sqrt{3}+2}{2\sqrt{3}+2}=2\sqrt{3}-\dfrac{\sqrt{3}(2\sqrt{3}+2)}{(2\sqrt{3}-2)(2\sqrt{3}+2)}=[/tex]
[tex]=2\sqrt{3}-\dfrac{2\cdot3+2\sqrt{3}}{(2\sqrt{3})^2-2^2}=2\sqrt{3}-\dfrac{6+2\sqrt{3} }{4\cdot3-4}=2\sqrt{3}-\dfrac{6+2\sqrt{3}}{12-4}=2\sqrt{3}-\dfrac{6+2\sqrt{3}}{8}=\\\\\\=2\sqrt{3}-\dfrac{\not2^1(3+\sqrt{3})}{\not8_{4}}=2\sqrt{3}-\dfrac{3+\sqrt{3}}{4}=\dfrac{4\cdot2\sqrt{3}}{4}-\dfrac{3+\sqrt{3}}{4}=\dfrac{8\sqrt{3}-(3+\sqrt{3})}{4}=\\\\=\dfrac{8\sqrt{3}-3-\sqrt{3}}{4}=\dfrac{7\sqrt{3}-3}{4}\\\\\\Odp.D[/tex]
