Korzystamy ze wzorów:
[tex]\sqrt[n]{a}\cdot\sqrt[n]{b}=\sqrt[n]{a\cdot b}\\\\\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{\dfrac{a}{b}}[/tex]
Zadanie 1
[tex]\dfrac{\sqrt[3]{112}}{\sqrt[3]{14}}=\sqrt[3]{\dfrac{112}{14}}=\sqrt[3]{8}=\sqrt[3]{2^3}=\boxed{2}[/tex]
Zadanie 2
[tex]\sqrt{18}\cdot\sqrt{50}=\sqrt{9\cdot2}\cdot\sqrt{25\cdot2}=\sqrt{9}\cdot\sqrt{2}\cdot\sqrt{25}\cdot\sqrt{2}=\\\\=3\cdot\sqrt{2}\cdot5\cdot\sqrt{2}=(3\cdot5)\cdot(\sqrt{2}\cdot\sqrt{2})=15\cdot2=\boxed{30}[/tex]
Zadanie 3
[tex]\sqrt[3]{36}\cdot\sqrt[3]{6}=\sqrt[3]{6^2}\cdot\sqrt[3]{6^1}=\sqrt[3]{6^2\cdot6^1}=\sqrt[3]{6^{2+1}}=\sqrt[3]{6^3}=\boxed{6}[/tex]
Zadanie 4
[tex]2\sqrt{3}\cdot(\sqrt{3}+2)=2\sqrt{3}\cdot\sqrt{3}+2\sqrt{3}\cdot2=2\cdot3+4\sqrt{3}=\boxed{6+4\sqrt{3}}[/tex]
Zadanie 5
[tex]3\sqrt{8}-5\sqrt{50}+\sqrt{2}=3\sqrt{4\cdot2}-5\sqrt{25\cdot2}+\sqrt{2}=\\\\=3\cdot\sqrt{4}\cdot\sqrt{2}-5\cdot\sqrt{25}\cdot\sqrt{2}+\sqrt{2}=\\\\=3\cdot2\cdot\sqrt{2}-5\cdot5\cdot\sqrt{2}+\sqrt{2}=\\\\=6\sqrt{2}-25\sqrt{2}+\sqrt{2}=\\\\=\boxed{-18\sqrt{2}}[/tex]
Zadanie 6
[tex]5\sqrt[3]{16}+3\sqrt[3]{54}-6\sqrt[3]{128}=5\sqrt[3]{8\cdot2}+3\sqrt[3]{27\cdot2}-6\sqrt[3]{64\cdot2}=\\\\=5\cdot\sqrt[3]{8}\cdot\sqrt[3]{2}+3\cdot\sqrt[3]{27}\cdot\sqrt[3]{2}-6\cdot\sqrt[3]{64}\cdot\sqrt[3]{2}=\\\\=5\cdot2\cdot\sqrt[3]{2}+3\cdot3\cdot\sqrt[3]{2}-6\cdot4\cdot\sqrt[3]{2}=\\\\=10\sqrt[3]{2}+9\sqrt[3]{2}-24\sqrt[3]{2}=\\\\=\boxed{-5\sqrt[3]{2}}[/tex]
