Odpowiedź:
Szczegółowe wyjaśnienie:
[tex]a)\sqrt{3}*\sqrt{\frac{5}{3}} =\sqrt{3}*\frac{\sqrt{5} }{\sqrt{3} }=\sqrt{5}\\b)\frac{6\sqrt{5} -4\sqrt{5} }{\sqrt{5} }=\frac{\sqrt{5}(6-4) }{\sqrt{5} }=6-4=2\\c)\frac{\sqrt{7}}{2\sqrt{7} +4\sqrt{7} }=\frac{\sqrt{7} }{\sqrt{7} (2+4)}=\frac{1}{6}\\d)\frac{8\sqrt{3} -2\sqrt{3} }{4\sqrt{3} -\sqrt{3} }=\frac{\sqrt{3} (8-2)}{\sqrt{3} (4-1)}=\frac{6}{3} =2\\e) 2\sqrt{3}*3\sqrt{2} =6\sqrt{3*2}=6\sqrt{6} \\f) \frac{6\sqrt{27} }{3\sqrt{3} } = 2\sqrt{\frac{27}{3} } =2\sqrt{9}=2*3=6\\[/tex]
[tex]g) \frac{\sqrt[3]{2} *\sqrt[3]{15} }{\sqrt[3]{6}}=\frac{\sqrt[3]{(2*15)}}{\sqrt[3]{6}} = \sqrt[3]{\frac{30}{6} }=\sqrt[3]{5} \\h\frac{2\sqrt[3]{3}*3\sqrt[3]{-10} }{\sqrt[3]{5} } =\frac{6\sqrt[3]{(3*(-10))} }{\sqrt[3]{5} } = 6\sqrt[3]{\frac{-30}{5}} =6\sqrt[3]{6}[/tex]
Podpowiedź:
[tex]\sqrt{a} * \sqrt{b} = \sqrt{(a*b)} \\\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b} }[/tex]
