W zadaniu należy uprościć podane wyrażenia. Skorzystamy z własności działań na pierwiastkach.

a)

$\sqrt{72}+\sqrt{32}+\sqrt{8}=\sqrt{36\cdot 2}+\sqrt{16\cdot 2}+\sqrt{4\cdot 2}=\underline{6}\sqrt{2}+\underline{4}\sqrt{2}+\underline{2}\sqrt{2}=\underline{12}\sqrt{2}$

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b)

$2\sqrt{75}-2\sqrt{3}+\sqrt{300}=2\sqrt{25\cdot 3}-2\sqrt{3}+\sqrt{100\cdot 3}=2\cdot 5\sqrt{3}-2\sqrt{3}+10\sqrt{3}=\underline{10}\sqrt{3}-\underline{2}\sqrt{3}+\underline{10}\sqrt{3}=\underline{18}\sqrt{3}$

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c)

$\sqrt[3]{54}+\sqrt[3]{128}-\sqrt[3]{432}=\sqrt[3]{27\cdot 2}+\sqrt[3]{64\cdot 2}-\sqrt[3]{216\cdot 2}=\sqrt[3]{3^3\cdot 2}+\sqrt[3]{4^3\cdot 2}-\sqrt[3]{6^3\cdot 2}=\underline{3}\sqrt[3]{2}+\underline{4}\sqrt[3]{2}-\underline{6}\sqrt[3]{2}=\underline{1}\sqrt[3]{2}=\sqrt[3]{2}$

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d)

$\sqrt[3]{2}\left(\sqrt[3]{500}-\sqrt[3]{4}\right)=\sqrt[3]{2\cdot 500}-\sqrt[3]{2\cdot 4}=\sqrt[3]{1000}-\sqrt[3]{8}=\sqrt[3]{10^3}-\sqrt[3]{2^3}=10-2=8$

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e)

$\frac{4\sqrt{3}+\sqrt{27}}{\sqrt{12}}=4\frac{\sqrt{3}}{\sqrt{12}}+\frac{\sqrt{27}}{\sqrt{12}}=4\sqrt{\frac{3}{12}}+\sqrt{\frac{27}{12}}=4\sqrt{\frac{1}{4}}+\sqrt{\frac{9}{4}}=4\cdot \frac{1}{2}+\frac{3}{2}=\frac{4}{2}+\frac{3}{2}=\frac{7}{2}=3,5$

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f)

$\frac{\sqrt{2}\left(\sqrt{6}+\sqrt{24}\right)}{\sqrt{3}}=\frac{\sqrt{2\cdot 6}+\sqrt{2\cdot 24}}{\sqrt{3}}=\frac{\sqrt{12}+\sqrt{48}}{\sqrt{3}}=\frac{\sqrt{12}}{\sqrt{3}}+\frac{\sqrt{48}}{\sqrt{3}}=\sqrt{\frac{12}{3}}+\sqrt{\frac{48}{3}}=\sqrt{4}+\sqrt{16}=2+4=6$