Odpowiedź:
[tex]a)\ \ \sqrt{32}-\sqrt{18}-\sqrt{8}=\sqrt{16\cdot2}-\sqrt{9\cdot2}-\sqrt{4\cdot2}=4\sqrt{2}-3\sqrt{2}-2\sqrt{2}=\\\\=\sqrt{2}-2\sqrt{2}=-\sqrt{2}\\\\\\b)\ \ \sqrt{20}+\sqrt{45}-\sqrt{125}=\sqrt{4\cdot5}+\sqrt{9\cdot5}-\sqrt{25\cdot5}=2\sqrt{5}+3\sqrt{5}-5\sqrt{5}=\\\\=5\sqrt{5}-5\sqrt{5}=0\\\\\\c)\ \ \sqrt{12}+\sqrt{27}-\sqrt{75}=\sqrt{4\cdot3}+\sqrt{9\cdot3}-\sqrt{25\cdot3}=2\sqrt{3}+3\sqrt{3}-5\sqrt{3}=\\\\=5\sqrt{3}-5\sqrt{3}=0[/tex]
[tex]d)\ \ \sqrt{160}-2\sqrt{90}+3\sqrt{40}=\sqrt{16\cdot10}-2\sqrt{9\cdot10}+3\sqrt{4\cdot10}=\\\\=4\sqrt{10}-2\cdot3\sqrt{10}+3\cdot2\sqrt{10}=4\sqrt{10}-6\sqrt{10}+6\sqrt{10}=4\sqrt{10}\\\\\\e)\ \ \sqrt[3]{48}+\sqrt[3]{162}=\sqrt[3]{8\cdot6}+\sqrt[3]{27\cdot6}=2\sqrt[3]{6}+3\sqrt[3]{6}=5\sqrt[3]{6}\\\\\\f)\ \ \sqrt[3]{135}-\sqrt[3]{40}=\sqrt[3]{27\cdot5}-\sqrt[3]{8\cdot5}=3\sqrt[3]{5}-2\sqrt[3]{5}=\sqrt[3]{5}[/tex]
