Answer: BCDA
Step-by-step explanation:
[tex]\dfrac{1}{\sqrt[4]{3x^3y^5}}\\\\\\\text{Since it is a 4th root, you need 4 on the "inside" to make 1 on the "outside"}\\=\dfrac{1}{y\sqrt[4]{3x^3y}}\\\\\\\text{Rationalize the denominator by multiplying so there are 4 on the "inside"}\\=\dfrac{1}{y\sqrt[4]{3x^3y}}\bigg(\dfrac{\sqrt[4]{3^3xy^3} }{\sqrt[4]{3^3xy^3}}\bigg)\\\\\\\text{Simplify}\\=\dfrac{\sqrt[4]{3^3xy^3} }{y(3xy)}\\\\\\\boxed{=\dfrac{\sqrt[4]{3^3xy^3} }{3xy^2}}\quad Option\ B[/tex]
[tex]\dfrac{3}{\sqrt[4]{3^3x^{11}y^{13}}}\\\\\\\text{Since it is a 4th root, you need 4 on the "inside" to make 1 on the "outside"}\\=\dfrac{3}{x^2y^3\sqrt[4]{3^3x^3y}}\\\\\\\text{Rationalize the denominator by multiplying so there are 4 on the "inside"}\\=\dfrac{3}{x^2y^3\sqrt[4]{3^3x^3y}}\bigg(\dfrac{\sqrt[4]{3xy^3} }{\sqrt[4]{3xy^3}}\bigg)\\\\\\\text{Simplify}\\=\dfrac{3\sqrt[4]{3xy^3} }{x^2y^3(3xy)}\\\\\\\boxed{=\dfrac{\sqrt[4]{3xy^3} }{x^3y^4}}\quad Option\ C[/tex]
[tex]\dfrac{2}{\sqrt[6]{2x^7y^5}}\\\\\\\text{Since it is a 6th root, you need 6 on the "inside" to make 1 on the "outside"}\\=\dfrac{2}{x\sqrt[6]{2xy^5}}\\\\\\\text{Rationalize the denominator by multiplying so there are 4 on the "inside"}\\=\dfrac{2}{x\sqrt[6]{2xy^5}}\bigg(\dfrac{\sqrt[6]{2^5x^5y} }{\sqrt[6]{2^5x^5y}}\bigg)\\\\\\\text{Simplify}\\=\dfrac{2\sqrt[6]{32x^5y} }{x(2xy)}\\\\\\\boxed{=\dfrac{\sqrt[6]{32x^5y} }{x^2y}}\quad Option\ D[/tex]
[tex]\dfrac{4}{\sqrt[6]{2^5x^5y^9}}\\\\\\\text{Since it is a 6th root, you need 6 on the "inside" to make 1 on the "outside"}\\=\dfrac{4}{y\sqrt[6]{2^5x^5y^3}}\\\\\\\text{Rationalize the denominator by multiplying so there are 4 on the "inside"}\\=\dfrac{4}{y\sqrt[6]{2^5x^5y^3}}\bigg(\dfrac{\sqrt[6]{2xy^3}}{\sqrt[6]{2xy^3}}\bigg)\\\\\\\text{Simplify}\\=\dfrac{4\sqrt[6]{2xy^3} }{y(2xy)}\\\\\\\boxed{=\dfrac{2\sqrt[6]{2xy^3} }{xy^2}}\quad Option\ A[/tex]
