Question 186669
In order to simplify {{{(1/9+1/(3x))/(x/9-1/x)}}} let's simplify the numerator and denominator separately first



Simplifying the numerator {{{1/9+1/(3x)}}}:




{{{1/9+1/(3x)}}} Start with the numerator. Take note that the LCD {{{9x/9x}}}




{{{x/(9x)+1/(3x)}}} Multiply the first fraction by {{{x/x}}}



{{{x/(9x)+3/(9x)}}} Multiply the second fraction by {{{3/3}}}



{{{(x+3)/(9x)}}} Add the fractions.



So {{{1/9+1/(3x)=(x+3)/(9x)}}} where {{{x<>0}}}



----------------------------------------------------------



Simplifying the denominator {{{x/9-1/x}}}: 



{{{x/9-1/x}}} Start with the denominator. Take note that the LCD {{{9x/9x}}}



{{{(x^2)/(9x)-1/x}}} Multiply the first fraction by {{{x/x}}}



{{{(x^2)/(9x)-9/(9x)}}} Multiply the second fraction by {{{9/9}}}



{{{(x^2-9)/(9x)}}} Subtract the fractions.



So {{{x/9-1/x=(x^2-9)/(9x)}}} where {{{x<>0}}}




---------------------------------------------------------------



Now that we've simplified the numerator of {{{(1/9+1/(3x))/(x/9-1/x)}}}, we then get:



{{{((x+3)/(9x))/((x^2-9)/(9x))}}}





{{{((x+3)/(9x))/((x^2-9)/(9x))}}} Start with the simplified expression.



{{{((x+3)/(9x))*((9x)/(x^2-9))}}} Multiply the first fraction by the reciprocal of the second fraction.



{{{((x+3)/(9x))*((9x)/((x+3)(x-3)))}}} Factor {{{x^2-9}}} to get {{{(x+3)(x-3)}}}



{{{(9x(x+3))/(9x(x+3)(x-3)))}}} Combine the fractions.



{{{(highlight(9x)highlight((x+3)))/(highlight(9x)highlight((x+3))(x-3)))}}} Highlight the common terms.



{{{(cross(9x)cross((x+3)))/(cross(9x)cross((x+3))(x-3)))}}} Cancel out the common terms.



{{{1/(x-3)}}} Simplify




==================================================================


Answer:


So {{{(1/9+1/(3x))/(x/9-1/x)}}} simplified to {{{((x+3)/(9x))/((x^2-9)/(9x))}}} which in turn simplified to {{{1/(x-3)}}}




So in short, {{{(1/9+1/(3x))/(x/9-1/x)}}} simplifies to {{{1/(x-3)}}}




In other words,  {{{(1/9+1/(3x))/(x/9-1/x)=1/(x-3)}}} where {{{x<>-3}}}, {{{x<>0}}}, or {{{x<>3}}}