Odpowiedź:
[tex]\dfrac{\sqrt{2}}{2\sqrt{5}}=\dfrac{\sqrt{5}}{5}\\\\\\\dfrac{\sqrt{2}}{2\sqrt{5}}\cdot\dfrac{\sqrt{5}}{\sqrt{5}}=\dfrac{\sqrt{5}}{5}\\\\\\\dfrac{\sqrt{10}}{2\cdot5}=\dfrac{\sqrt{5}}{5}\\\\\\\dfrac{\sqrt{10}}{10}=\dfrac{\sqrt{5}}{5}\\\\L\neq P\\\\NIE[/tex]
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[tex]\dfrac{\sqrt{5}-1}{2\sqrt{5}}=\dfrac{5-\sqrt{5}}{10}\\\\\\\dfrac{\sqrt{5}-1}{2\sqrt{5}}\cdot\dfrac{\sqrt{5}}{\sqrt{5}}=\dfrac{5-\sqrt{5}}{10}\\\\\\\dfrac{(\sqrt{5}-1)\cdot\sqrt{5} }{2\cdot5}=\dfrac{5-\sqrt{5}}{10}\\\\\\\dfrac{5-\sqrt{5}}{10}=\dfrac{5-\sqrt{5}}{10}\\\\L=P\\\\TAK[/tex]
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[tex]\dfrac{2\sqrt{5}}{\sqrt{5}-2}=10+4\sqrt{5}\\\\\\\dfrac{2\sqrt{5}}{\sqrt{5}-2}\cdot\dfrac{\sqrt{5}+2}{\sqrt{5}+2}=10+4\sqrt{5}\\\\\\\dfrac{2\sqrt{5}(\sqrt{5}+2)}{(\sqrt{5}-2)(\sqrt{5}+2)}=10+4\sqrt{5}\\\\\\\dfrac{2\cdot5+4\sqrt{5}}{(\sqrt{5})^2-2^2}=10+4\sqrt{5}\\\\\\\dfrac{10+4\sqrt{5}}{5-4}=10+4\sqrt{5}\\\\\\\dfrac{10+4\sqrt{5}}{1}=10+4\sqrt{5}\\\\\\10+4\sqrt{5}=10+4\sqrt{5}\\\\L=P\\\\TAK[/tex]
