wyłącz czynnik przed znak pierwiastku

[tex]\sqrt{12}=\sqrt{4\cdot3}=\sqrt{2^2\cdot3}=2\sqrt3\\\\\sqrt{28}=\sqrt{4\cdot7}=\sqrt{2^2\cdot7}=2\sqrt7\\\\\sqrt{80}=\sqrt{16\cdot5}=\sqrt{4^2\cdot5}=4\sqrt5\\\\\sqrt{125}=\sqrt{25\cdot5}=\sqrt{5^2\cdot5}=5\sqrt5[/tex]

Od e) do h)

[tex]\sqrt[3]{32}=\sqrt[3]{8\cdot4}=\sqrt[3]{2^3\cdot4}=2\sqrt[3]4\\\\\sqrt[3]{54}=\sqrt[3]{27\cdot2}=\sqrt[3]{3^3\cdot2}=3\sqrt[3]2\\\\\sqrt[3]{-16}=\sqrt[3]{-8\cdot2}=\sqrt[3]{(-2)^3\cdot2}=-2\sqrt[3]2\\\\\sqrt[3]{-128}=\sqrt[3]{-64\cdot2}=\sqrt[3]{(-4)^3\cdot2}=-4\sqrt[3]2[/tex]

TABIASZCZON TABIASZCZON · Answer 2

A] [tex]\sqrt{12}=\sqrt{4\cdot3}=\sqrt{4} \cdot\sqrt{3}=2\sqrt{3}[/tex]

B] [tex]\sqrt{28}= \sqrt{4\cdot7}= \sqrt{4}\cdot \sqrt{7} =2\sqrt{7}[/tex]

C] [tex]\sqrt{80}= \sqrt{16\cdot5}= \sqrt{16}\cdot \sqrt{5}=4 \sqrt{5}[/tex]

D] [tex]\sqrt{125}= \sqrt{25\cdot5} =\sqrt{25}\cdot \sqrt{3}=5 \sqrt{3}[/tex]

E] [tex]\sqrt[3]{32}= \sqrt[3]{8\cdot4}= \sqrt[3]{8}\cdot \sqrt[3]{4} =2\sqrt[3]{4}[/tex]

F] [tex]\sqrt[3]{54}= \sqrt[3]{27\cdot2} =\sqrt[3]{27} \cdot\sqrt[3]{2}=3 \sqrt[3]{2}[/tex]

G] [tex]\sqrt[3]{-16}= \sqrt[3]{-8\cdot2}= \sqrt[3]{-8}\cdot \sqrt[3]{2}=-2 \sqrt[3]{2}[/tex]

H] [tex]\sqrt[3]{-128}= \sqrt[3]{-64\cdot2}= \sqrt[3]{-64}\cdot \sqrt[3]{2}=-4 \sqrt[3]{2}[/tex]