Question 722693
<pre>
{{{((x-1)/(x^2-4))/(1+1/(x-2))}}}{{{""-""}}}{{{1/(x-2)}}}

Do the left numerator by factoring the denominator:

               {{{(x-1)/(x^2-4)}}} = {{{(x-1)/((x-2)(x+2))}}}

Do the left denominator by getting an LCD of (x-2):

               {{{1+1/(x-2)}}} = {{{(x-2)/(x-2)}}}{{{""+""}}}{{{1/(x-2)}}} = {{{(x-2+1)/(x-2)}}} = {{{(x-1)/(x-2)}}}

So we have

{{{((x-1)/((x-2)(x+2)))/((x-1)/(x-2))}}}{{{""-""}}}{{{1/(x-2)}}}

The first term is a fraction divided by a fraction, so

{{{(x-1)/((x-2)(x+2))}}}{{{"÷"}}}{{{(x-1)/(x-2)}}}{{{""-""}}}{{{1/(x-2)}}}

So we invert the second fraction and change the division to 
multiplication.

{{{(x-1)/((x-2)(x+2))}}}{{{"*"}}}{{{(x-2)/(x-1)}}}{{{""-""}}}{{{1/(x-2)}}}

We cancel the (x-2)'s

{{{(x-1)/((cross(x-2))(x+2))}}}{{{"*"}}}{{{(cross(x-2))/(x-1)}}}{{{""-""}}}{{{1/(x-2)}}}

We cancel the (x-1)'s

{{{(""^1cross(x-1))/((cross(x-2))(x+2))}}}{{{"*"}}}{{{(cross(x-2))/(cross(x-1))}}}{{{""-""}}}{{{1/(x-2)}}}


{{{1/(x+2)}}}{{{""-""}}}{{{1/(x-2)}}}

Get an LCD of (x+2)(x-2)

{{{(1*red((x-2)))/((x+2)*red((x-2)))}}}{{{""-""}}}{{{(1*red((x+2)))/((x-2)*red((x+2)))}}}


{{{((x-2)-(x+2))/((x+2)*(x-2))}}}

{{{(x-2-x-2)/((x+2)*(x-2))}}}

{{{(cross(x)-2-cross(x)-2)/((x+2)*(x-2))}}}

{{{(-2-2)/((x+2)*(x-2))}}}

{{{(-4)/((x+2)*(x-2))}}}

So the answer has -4 in the numerator, not +4.

Edwin</pre>