Odpowiedź:
[tex]2\sqrt{32}+4\sqrt{2}=2\sqrt{16\cdot2}+4\sqrt{2}=2\cdot4\sqrt{2}+4\sqrt{2}=8\sqrt{2}+4\sqrt{2}=12\sqrt{2}\\\\\\\sqrt{45}-\sqrt{5}=\sqrt{9\cdot5}-\sqrt{5}=3\sqrt{5}-\sqrt{5}=2\sqrt{5}\\\\\\2\sqrt{27}+\sqrt{12}=2\sqrt{9\cdot3}+\sqrt{4\cdot3}=2\cdot3\sqrt{3}+2\sqrt{3}=6\sqrt{3}+2\sqrt{3}=8\sqrt{3}\\\\\\7\sqrt{32}-2\sqrt{18}=7\sqrt{16\cdot2}-2\sqrt{9\cdot2}=7\cdot4\sqrt{2}-2\cdot3\sqrt{2}=28\sqrt{2}-6\sqrt{2}=22\sqrt{2}[/tex]
[tex]2\sqrt[3]{81}+\sqrt[3]{24}=2\sqrt[3]{27\cdot3}+\sqrt[3]{8\cdot3}=2\cdot3\sqrt[3]{3} +2\sqrt[3]{3}=6\sqrt[3]{3}+2\sqrt[3]{3}=8\sqrt[3]{3}\\\\\\7\sqrt[3]{16}-2\sqrt[3]{54}=7\sqrt[3]{8\cdot2}-2\sqrt[3]{27\cdot2}=7\cdot2\sqrt[3]{2}-2\cdot3\sqrt[3]{2}=14\sqrt[3]{2}-6\sqrt[3]{2}=8\sqrt[3]{2}[/tex]
