Start by factoring everything:
#=(9x^2 + 3x - 21x - 7)/(2(x - 2)) xx (x - 2)/((3x - 7)(3x + 7))#
#=(3x(3x + 1) - 7(3x + 1))/(2(x - 2)) xx (x - 2)/((3x - 7)(3x + 7))#
#=((3x - 7)(3x + 1))/(2(x - 2)) xx (x - 2)/((3x - 7)(3x + 7))#
Eliminate what you can:
#=(cancel(3x - 7)(3x + 1))/(2(cancel(x - 2))) xx (cancel(x - 2))/(cancel(3x - 7)(3x + 7))#
#=(3x + 1)/(2(3x + 7))#
Note the restrictions on the variable. These are found by setting the denominator to #0# and solving in the original expression.
#2x - 4 = 0#
#2x = 4#
#x = 2#
AND
#9x ^2 - 49 = 0#
#(3x + 7)(3x - 7) = 0#
#x = -7/3 and 7/3#
Hence, #(9x^2 - 18x - 7)/(2x - 4) xx (x - 2)/(9x^2 - 49) = (3x + 1)/(2(3x + 7)), x != 2, 7/3, -7/3#
Hopefully this helps!
