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Uporzadkuj rosnąco liczby:
\frac{1}{2}\sqrt{184}=\sqrt{\frac{184}{4}}=\sqrt{46}
4\sqrt3=\sqrt{4^2*3}=\sqrt{48}
\sqrt{50}
3\sqrt6=\sqrt{3^2*6}=\sqrt{54}
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-\frac{10\sqrt[3]{-12}}{2}-4(\sqrt[3]{12}-5)=-5*(-\sqrt[3]{12})-4\sqrt[3]{12}+20=5\sqrt[3]{12}-4\sqrt[3]{12}+20=20+\sqrt[3]{12}
b)
\sqrt[3]{3^3}-\frac{1}{5}\sqrt{75}+2\sqrt{48}=3-\frac{1}{5}*5\sqrt3+2*4\sqrt3=3-\sqrt3+8\sqrt3=3+7\sqrt3
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\frac{3\sqrt2+5}{2\sqrt3}=\frac{(3\sqrt2+5)*\sqrt3}{2\sqrt3*\sqrt3}=\frac{3\sqrt6+5\sqrt3}{2*3}=\frac{3\sqrt6+5\sqrt3}{6}
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\frac{\frac{\sqrt{3^4}}{3}-(\frac{\sqrt3}{2})^{-2}}{(\frac{2}{\sqrt3})^2*(\frac{1}{\sqrt3})^{-3}}=\frac{\frac{\sqrt{81}}{3}-(\frac{2}{\sqrt3})^{2}}{\frac{4}{(\sqrt3)^2}*\frac{(\sqrt3)^3}{1^3}}=
\frac{\frac{9}{3}-\frac{4}{3}}{4*\sqrt3}=\frac{5}{3}*\frac{1}{4\sqrt3}=\frac{5}{12\sqrt3}=
\frac{5\sqrt3}{12\sqrt3*\sqrt3}=\frac{5\sqrt3}{36}