Answer:
[tex]\frac{\sqrt[4]{3x^2} }{2y}[/tex]
Step-by-step explanation:
We can simplify the expression under the root first.
Remember to use [tex]\frac{a^x}{a^y}=a^{x-y}[/tex]
Thus, we have:
[tex]\sqrt[4]{\frac{24x^{6}y}{128x^{4}y^{5}}} \\=\sqrt[4]{\frac{3x^{2}}{16y^{4}}}[/tex]
We know 4th root can be written as "to the power 1/4th". Then we can use the property [tex](ab)^{x}=a^x b^x[/tex]
So we have:
[tex]\sqrt[4]{\frac{3x^{2}}{16y^{4}}} \\=(\frac{3x^{2}}{16y^{4}})^{\frac{1}{4}}\\=\frac{3^{\frac{1}{4}}x^{\frac{1}{2}}}{2y}\\=\frac{\sqrt[4]{3x^2} }{2y}[/tex]
Option D is right.
