Szczegółowe wyjaśnienie:
[tex]\sqrt[3]{81}:\sqrt[3]3=\sqrt[3]{81:3}=\sqrt[3]{27}=3\ \text{bo}\ 3^3=27\\\\\sqrt[3]4\cdot\sqrt[3]{16}=\sqrt[3]{4\cdot16}=\sqrt[3]{64}=4\ \text{bo}\ 4^3=64\\\\\sqrt[3]{216:(-27)}=\sqrt[3]{216}\cdot\sqrt[3]{-27}=6\cdot(-3)=-18\\\\\sqrt[3]{3,2}:\sqrt[3]{0,4}=\sqrt[3]{3,2:0,4}=\sqrt[3]{32:4}=\sqrt[3]8=2[/tex]
Skorzystałem z twierdzeń:
[tex]\sqrt[3]{a\cdot b}=\sqrt[3]{a}\cdot\sqrt[3]{b}\\\\\sqrt[3]{a:b}=\sqrt[3]{a}:\sqrt[3]{b}\ \text{gdzie}\ b\neq0[/tex]
[tex]\sqrt{16}+\sqrt9-\sqrt{16+9}=4+3-\sqrt{25}=7-5=2\\\\\sqrt{16}=4\ \text{bo}\ 4^2=16\\\sqrt9=3\ \text{bo}\ 3^2=9\\\sqrt{25}=5\ \text{bo}\ 5^2=25[/tex]
[tex]\sqrt{36+64}=\sqrt{100}=10\ \text{bo}\ 10^2=100[/tex]
[tex]4\sqrt[3]{-8}-\sqrt{64}=4\cdot(-2)-8=-8-8=-16\\\\\sqrt[3]{-8}=-2\ \text{bo}\ (-2)^3=-8\\\sqrt{64}=8\ \text{bo}\ 8^2=64[/tex]
[tex]\sqrt[3]{9^2-17}-\sqrt{4^3}=\sqrt[3]{81-17}-\sqrt{64}=\sqrt[3]{64}-8=4-8=-4\\\\\sqrt[3]{64}=4\ \text{bo}\ 4^3=64\\\sqrt{64}=8\ \text{bo}\ 8^2=64[/tex]
