Odpowiedź:
[tex]a)\ \ -\dfrac{\not10^5\sqrt[3]{-12}}{\not2_{1}}-4(\sqrt[3]{12}-5)=-5\cdot\sqrt[3]{-12}-4\sqrt[3]{12}+20=\\\\=-5\cdot(-\sqrt[3]{12})-4\sqrt[3]{12}+20=5\sqrt[3]{12}-4\sqrt[3]{12}+20=\sqrt[3]{12}+20\\\\\\b)\ \ \sqrt{3^3}-\frac{1}{5}\sqrt{75}+2\sqrt{48}=\sqrt{27}-\frac{1}{5}\sqrt{25\cdot3}+2\sqrt{16\cdot3}=\sqrt{9\cdot3}-\frac{1}{\not5}\cdot\not5\sqrt{3}+2\cdot4\sqrt{3}=\\\\=3\sqrt{3}-\sqrt{3}+8\sqrt{3}=2\sqrt{3}+8\sqrt{3}=10\sqrt{3}[/tex]
8.
a) = - 5(3√-12) - 4(3√12) + 20) = 3√12 + 20
b) = 3√3 - 1/3 (√25 * √3) + 2 (√16 * √3) = 3√3 - 1/5(5√3) + 2(4√3) = 3√3 - 3√3 + 8√3 = 10√3
