Answer:
See explanation
Step-by-step explanation:
Use facts
[tex]a=\sqrt{a^2}\ [\text{for all} \ a\ge 0]\\ \\a=\sqrt[n]{a^n}\ [\text{for all} \ a\ge 0][/tex]
Hence,
[tex]2\sqrt{2}=\sqrt{4}\cdot \sqrt{2}=\sqrt{4\cdot 2}=\sqrt{8}\\ \\3\sqrt{2}=\sqrt{9}\cdot \sqrt{2}=\sqrt{9\cdot 2}=\sqrt{18}\\ \\5\sqrt{3}=\sqrt{25}\cdot \sqrt{3}=\sqrt{25\cdot 3}=\sqrt{75}\\ \\\dfrac{1}{3}\sqrt{6}=\sqrt{\dfrac{1}{9}}\cdot \sqrt{6}=\sqrt{\dfrac{1}{9}\cdot 6}=\sqrt{\dfrac{2}{3}}\\ \\\dfrac{3}{5}\sqrt{5}=\sqrt{\dfrac{9}{25}}\cdot \sqrt{5}=\sqrt{\dfrac{9}{25}\cdot 5}=\sqrt{\dfrac{9}{5}}\\ \\3\sqrt[3]{3}=\sqrt[3]{27}\cdot \sqrt[3]{3}=\sqrt[3]{27\cdot 3}=\sqrt[3]{81}[/tex]
[tex]8\sqrt{2}=\sqrt{64}\cdot \sqrt{2}=\sqrt{64\cdot 2}=\sqrt{128}\\ \\5\sqrt[5]{3}=\sqrt[5]{5^5}\cdot \sqrt[5]{3}=\sqrt[5]{5^5\cdot 3}=\sqrt[5]{9,375}\\ \\\dfrac{4}{3}\sqrt{3}=\sqrt{\dfrac{16}{9}}\cdot \sqrt{3}=\sqrt{\dfrac{16}{9}\cdot 3}=\sqrt{\dfrac{16}{3}}\\ \\2\sqrt{\dfrac{1}{2}}=\sqrt{4}\cdot \sqrt{\dfrac{1}{2}}=\sqrt{4\cdot \dfrac{1}{2}}=\sqrt{2}[/tex]
