Question

Can somebody please explain how to get this answer? x^(2/3)y^(-1/4)/(x^(1/2)y^(-1/2)

Answer 1

repeated use of the principle that x^a / x^b = x^(a-b)

Answer 2

\[\huge \frac {x^{2/3}y^{-1/4}} {x^{1/2}y^{-1/2}}\]
\[\huge x^{2/3}y^{-1/4} ~~~\times ~~~ (x^{1/2}y^{-1/2})^{-1} \]
\[\huge x^{2/3}y^{-1/4} ~~~\times ~~~ x^{-1/2}y^{1/2} \]
\[\huge x^{2/3}\times y^{-1/4} ~~~\times ~~~ x^{-1/2} \times y^{1/2} \]
\[\huge x^{2/3}\times x^{-1/2} \times y^{-1/4} \times y^{1/2} \]
\[\huge x^{(2/3)+(-1/2)} \times y^{(-1/4)+(1/2)} \]
\[\huge x^{(4/6)+(-3/6)} \times y^{(-2/8)+(4/8)} \]
\[\huge x^{(1/6)} y^{(2/8)}~~ =~~ x^{(1/6)} y^{(1/4)} \]
\[\huge \sqrt[6]{x}\times \sqrt[4]{y}\]

Answer 3

Thank you so much!!!

Answer 4

yw @Misskatierae