Question 269084
Remember that {{{(x^(y))/(x^(z))=x^(y-z)}}}



So this means that {{{(x^(1/2)^"")/(x^(1/3)^"")=x^(1/2-1/3)=x^(3/6-2/6)=x^((3-2)/6)=x^(1/6)^""}}}



So {{{(x^(1/2)^"")/(x^(1/3)^"")=x^(1/6)^""}}} where {{{x<>0}}}