which expression is equivalent to $\frac{sqrt4{6}}{sqrt3{2}}$?\n$\frac{sqrt12{27}}{2}$\n$\frac{sqrt4{24}}{2}$…

Explanation:

Step1: Rationalize the denominator

Multiply the numerator and denominator by $\sqrt[3]{2^{2}}$ to get a rational - denominator. The original expression $\frac{\sqrt[4]{6}}{\sqrt[3]{2}}$ becomes $\frac{\sqrt[4]{6}\times\sqrt[3]{2^{2}}}{\sqrt[3]{2}\times\sqrt[3]{2^{2}}}=\frac{\sqrt[4]{6}\times\sqrt[3]{4}}{2}$.

Step2: Rewrite the radicals with a common index

The least - common multiple of 4 and 3 is 12. $\sqrt[4]{6}=6^{\frac{1}{4}} = 6^{\frac{3}{12}}=\sqrt[12]{6^{3}}=\sqrt[12]{216}$, and $\sqrt[3]{4}=4^{\frac{1}{3}}=4^{\frac{4}{12}}=\sqrt[12]{4^{4}}=\sqrt[12]{256}$. Then $\frac{\sqrt[4]{6}\times\sqrt[3]{4}}{2}=\frac{\sqrt[12]{216}\times\sqrt[12]{256}}{2}$.

Step3: Use the property $\sqrt[n]{a}\times\sqrt[n]{b}=\sqrt[n]{ab}$

$\frac{\sqrt[12]{216}\times\sqrt[12]{256}}{2}=\frac{\sqrt[12]{216\times256}}{2}=\frac{\sqrt[12]{55296}}{2}$.

Answer:

$\frac{\sqrt[12]{55296}}{2}$ (the third option)