Zad.10/80. Matematyka. Pierwiastki. Podpunkt e,f,g,h
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[tex]e)\\\\ \sqrt[3]{54}-2\sqrt[3]{16}=\sqrt[3]{27*2}-2\sqrt[3]{8*2}=\sqrt[3]{27 }*\sqrt[3]{2}-2\sqrt[3]{8 }*\sqrt[3]{2}=3\sqrt[3]{2}-2*2\sqrt[3]{2}=\\\\=3\sqrt[3]{2}-4\sqrt[3]{2}=-\sqrt[3]{2}\\\\f)\\\\-\sqrt[3] {40}+\sqrt[3]{320}=-\sqrt[3] {8*5}+\sqrt[3]{64*5}=-\sqrt[3] {8 }*\sqrt[3]{5}+\sqrt[3]{64}*\sqrt[3]{5}=\\\\=-2 \sqrt[3]{5}+4 \sqrt[3]{5}=2*\sqrt[3]{5}[/tex]
[tex]g)\\\\5\sqrt{12}+4\sqrt{75}-3\sqrt{48}=5\sqrt{4*3}+4\sqrt{25*3}-3\sqrt{16*3}=\\\\=5\sqrt{4 }*\sqrt{3}+4\sqrt{25 }*\sqrt{3}-3\sqrt{16 }*\sqrt{3}=5*2 *\sqrt{3}+4*5*\sqrt{3}-3*4*\sqrt{3}=\\\\=10\sqrt{3}+20\sqrt{3}-12\sqrt{3}=18\sqrt{3} \\\\h)\\\\6\sqrt[3]{16}+7\sqrt[3]{54}-5\sqrt[3]{128} =6\sqrt[3]{8*2}+7\sqrt[3]{27*2}-5\sqrt[3]{64*2}=\\\\ =6\sqrt[3]{8 }*\sqrt[3]{2}+7\sqrt[3]{27 }*\sqrt[3]{2}-5\sqrt[3]{64 }*\sqrt[3]{2}=6*2 *\sqrt[3]{2}+7*3 *\sqrt[3]{2}-5*4 *\sqrt[3]{2}=\\\\=12 \sqrt[3]{2}+21*\sqrt[3]{2}-20 \sqrt[3]{2}=13\sqrt[3]{2}[/tex]