3.
[tex]a) \ \sqrt{72}+\sqrt{32}+\sqrt{8} = \sqrt{36\cdot2}+\sqrt{16\cdot2}+\sqrt{4\cdot2}=\sqrt{36}\cdot\sqrt{2}+\sqrt{16}\cdot\sqrt{2}+\sqrt{4}\cdot\sqrt{2} =\\\\=6\cdot\sqrt{2}+4\cdot\sqrt{2}+2\cdot\sqrt{2} = 6\sqrt{2}+4\sqrt{2}+2\sqrt{2} = \boxed{12\sqrt{2}}[/tex]
[tex]c) \ \sqrt[3]{54}+\sqrt[3]{128}-\sqrt[3]{432} = \sqrt[3]{27\cdot2}+\sqrt[3]{64\cdot2}-\sqrt[3]{216\cdot2}=\\\\=\sqrt[3]{27}\cdot\sqrt[3]{2}+\sqrt[3]{64}\cdot\sqrt[3]{2}-\sqrt[3]{216}\cdot\sqrt[3]{2}=3\cdot\sqrt[3]{2}+4\cdot\sqrt[3]{2}-6\cdot\sqrt[3]{2} =3\sqrt[3]{2}+4\sqrt[3]{2}-6\sqrt[3]{2}=\\\\=\boxed{\sqrt{2}}[/tex]
[tex]e) \ \frac{4\sqrt{3}+\sqrt{27}}{\sqrt{12}} = \frac{4\sqrt{3}+\sqrt{9\cdot3}}{\sqrt{4\cdot3}} = \frac{4\sqrt{3}+\sqrt{9}\cdot\sqrt{3}}{\sqrt{4}\cdot\sqrt{3}} = \frac{4\sqrt{3}+3\sqrt{3}}{2\sqrt{3}} = \frac{7\sqrt{3}}{2\sqrt{3}} = \frac{7}{2} = \boxed{3,5}[/tex]
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Verified answer
3.
[tex]a) \ \sqrt{72}+\sqrt{32}+\sqrt{8} = \sqrt{36\cdot2}+\sqrt{16\cdot2}+\sqrt{4\cdot2}=\sqrt{36}\cdot\sqrt{2}+\sqrt{16}\cdot\sqrt{2}+\sqrt{4}\cdot\sqrt{2} =\\\\=6\cdot\sqrt{2}+4\cdot\sqrt{2}+2\cdot\sqrt{2} = 6\sqrt{2}+4\sqrt{2}+2\sqrt{2} = \boxed{12\sqrt{2}}[/tex]
[tex]c) \ \sqrt[3]{54}+\sqrt[3]{128}-\sqrt[3]{432} = \sqrt[3]{27\cdot2}+\sqrt[3]{64\cdot2}-\sqrt[3]{216\cdot2}=\\\\=\sqrt[3]{27}\cdot\sqrt[3]{2}+\sqrt[3]{64}\cdot\sqrt[3]{2}-\sqrt[3]{216}\cdot\sqrt[3]{2}=3\cdot\sqrt[3]{2}+4\cdot\sqrt[3]{2}-6\cdot\sqrt[3]{2} =3\sqrt[3]{2}+4\sqrt[3]{2}-6\sqrt[3]{2}=\\\\=\boxed{\sqrt{2}}[/tex]
[tex]e) \ \frac{4\sqrt{3}+\sqrt{27}}{\sqrt{12}} = \frac{4\sqrt{3}+\sqrt{9\cdot3}}{\sqrt{4\cdot3}} = \frac{4\sqrt{3}+\sqrt{9}\cdot\sqrt{3}}{\sqrt{4}\cdot\sqrt{3}} = \frac{4\sqrt{3}+3\sqrt{3}}{2\sqrt{3}} = \frac{7\sqrt{3}}{2\sqrt{3}} = \frac{7}{2} = \boxed{3,5}[/tex]