#(x^2-4)/(12x) div (2-x)/(4xy)#
Whenever we have a complex division, may it is simpler to turn it into a mutiplication #a div (b/c)= a xx (c/b)#:
#(x^2-4)/(12x) xx (4xy)/(2-x)#
We can now exchange the denominators, because multiplication is permutable:
#(x^2-4)/(2-x) xx (4xy)/(12x)#
Let's turn #2-x# in a expression that begins by #x#. Doesn't have any effect, but I need it to develope the reasoning:
#(x^2-4)/(-x+2) xx (4xy)/(12x)#
Now, let's take the minus sign of x to outside of the expression:
#-(x^2-4)/(x-2) xx (4xy)/(12x)#
#x^2-4# is on the form #a^2-b^2#, which is (a+b)(a-b):
#-((x-2)(x+2))/(x-2) xx (4xy)/(12x)#
Now we can cut the factors in common between numerators and denominators:
#-(cancel(x-2)(x+2))/cancel(x-2) xx (4cancel(x)y)/(12cancel(x))#
#-(x+2) xx (4y)/(12)#
Now, you only need to divide 12 by 4:
#-(x+2) y /(3)#
