Zadanie 1
Korzystamy ze wzoru:
[tex](a+b)(a-b)=a^2-b^2[/tex]
Mamy:
[tex]\dfrac{15}{4-\sqrt{6}}=\dfrac{15\cdot(4+\sqrt{6})}{(4-\sqrt{6})(4+\sqrt{6})}=\dfrac{15\cdot(4+\sqrt{6})}{4^2-(\sqrt{6})^2}=\\\\\\=\dfrac{15\cdot(4+\sqrt{6})}{16-6}=\dfrac{15\cdot(4+\sqrt{6})}{10}=\dfrac{3\cdot(4+\sqrt{6})}{2}=\\\\\\=\boxed{\dfrac{12+3\sqrt{6}}{2}}[/tex]
Zadanie 2
Korzystamy z tego, że:
[tex]\sqrt[3]{2}\cdot\sqrt[3]{2}\cdot\sqrt[3]{2}=\sqrt[3]{2^3}=2[/tex]
Mamy:
[tex]\dfrac{8}{\sqrt[3]{2}}=\dfrac{8\cdot\sqrt[3]{2}\cdot\sqrt[3]{2}}{\sqrt[3]{2}\cdot\sqrt[3]{2}\cdot\sqrt[3]{2}}=\dfrac{8\sqrt[3]{4}}{2}=\boxed{4\sqrt[3]{4}}[/tex]
