Question 57355
If this is what you meant:
{{{x/(x-4)-1/(x-1)}}} divided by {{{x/(x-1)+2/(x-3)}}}
This is how you do it:  It's probably easiest to turn this into two fractions then divide rather than using the methods commonly used for complex fractions.
{{{x/(x-4)-1/(x-1)}}} divided by {{{x/(x-1)+2/(x-3)}}}
:
The LCD for the 1st fraction is: (x-4)(x-1) whatever you do to the denominator to get that, you also do to the numerator:
{{{x(x-1)/((x-4)(x-1))-1(x-4)/((x-4)(x-1))}}}
{{{(x^2-x)/((x-4)(x-1))+(-x+4)/((x-4)(x-1))}}}
{{{(x^2-x-x+4)/((x-4)(x-1))}}}
{{{(x^2-2x+4)/((x-4)(x-1))}}}  
:
The LCD for the 2nd fraction is: (x-1)(x-3)
{{{x(x-3)/((x-1)(x-3))+2(x-1)/((x-1)(x-3))}}}
{{{(x(x-3)+2(x-1))/((x-1)(x-3))}}}
{{{(x^2-3x+2x-2)/((x-1)(x-3))}}}
{{{(x^2-x-2)/((x-1)(x-3))}}} 
:
Now you have:
{{{(x^2-2x+4)/((x-4)(x-1))}}} divided by {{{(x^2-x-2)/((x-1)(x-3))}}}  Flip the second fraction and multiply:
{{{((x^2-2x+4)/((x-4)(x-1)))*(((x-1)(x-3))/(x^2-x-2))}}}
{{{((x^2-2x+4)/((x-4)*cross((x-1))))*((cross((x-1))*(x-3))/(x^2-x-2))}}}
{{{((x^2-2x+4)/(x-4))*((x-3)/(x^2-x-2))}}}
{{{(x-3)(x^2-2x+4)/((x-4)(x^2-x-2))}}}
Happy Calculating, that was a whopper!!!